更多答案也可以关注我的GitHub算法仓库——Algorithm

（大家可以根据我每道题打的星来衡量这道题的难易，星越少越难）
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所有代码都是在力扣上提交通过的，保证代码的正确无误

基础数据结构：

class ListNode {
        int val;
        ListNode next;
        ListNode() {}
        ListNode(int val) { this.val = val; }
        ListNode(int val, ListNode next) { this.val = val; this.next = next; }
    }
class TreeNode {
       int val;
       TreeNode left;
       TreeNode right;
       TreeNode() {}
       TreeNode(int val) { this.val = val; }
       TreeNode(int val, TreeNode left, TreeNode right) {
           this.val = val;
           this.left = left;
           this.right = right;
       }
    }

92.反转链表II

思路是记录反转部分的前一个节点和后一个节点，将其反转后再连接，思路是这个，但是我在实现上可能麻烦一些：

class Solution {
    public ListNode reverseBetween(ListNode head, int left, int right) {
        ListNode curr = head;
        ListNode prev = null;
        ListNode next = curr;
        int flag = 1;
        while (flag < right) {
            if (flag == left - 1) {
                prev = curr;
            }
            curr = curr.next;
            flag++;
        }
        next = curr.next;
        if (prev == null) {
            head = reverse(head, left, right);
            prev = head;
        }else {
            prev.next = reverse(prev.next, left, right);
        }
        while ( prev.next != null && left <= right) {
            prev = prev.next;
        }
        if (next != null) {
            if (prev.next != null) {
                prev.next.next = next;
            }else {
                prev.next = next;
            }
        }
        return head;
    }

    public ListNode reverse(ListNode head, int left, int right) {
        if (left == right) {
            return head;
        }
        ListNode prev = null;
        ListNode curr = head;
        while (left <= right) {
            ListNode nextTemp = curr.next;
            curr.next = prev;
            prev = curr;
            curr = nextTemp;
            left++;
        }
        return prev;
    }
}

200.岛屿数量

广搜、深搜都可以，这里是深搜+剪枝，不用回溯：

class Solution {
    public int numIslands(char[][] grid) {
        int res = 0;
        for (int i = 0; i < grid.length; i++) {
            for (int j = 0; j < grid[0].length; j++) {
                if (grid[i][j] == '1') {
                    res++;
                    dfs(grid, i, j);
                }
            }
        }
        return res;
    }
    public void dfs(char[][] grid, int x, int y) {
        grid[x][y] = '0';
        if (x < grid.length - 1 && grid[x+1][y] == '1') {
            dfs(grid, x+1, y);
        }
        if (x > 0 && grid[x-1][y] == '1') {
            dfs(grid, x-1, y);
        }
        if (y < grid[0].length - 1 && grid[x][y+1] == '1') {
            dfs(grid, x, y+1);
        }
        if (y > 0 && grid[x][y-1] == '1') {
            dfs(grid, x, y-1);
        }
    }
}

33.搜索旋转排序数组

二分（我当时是懒省事直接O(n)遍历了）

class Solution {
    public int search(int[] nums, int target) {
        int n = nums.length;
        if (n == 0) {
            return -1;
        }
        if (n == 1) {
            return nums[0] == target ? 0 : -1;
        }
        int l = 0, r = n - 1;
        while (l <= r) {
            int mid = (l + r) / 2;
            if (nums[mid] == target) {
                return mid;
            }
            if (nums[0] <= nums[mid]) {
                if (nums[0] <= target && target < nums[mid]) {
                    r = mid - 1;
                } else {
                    l = mid + 1;
                }
            } else {
                if (nums[mid] < target && target <= nums[n - 1]) {
                    l = mid + 1;
                } else {
                    r = mid - 1;
                }
            }
        }
        return -1;
    }
}

46.全排列

经典算法，使用dfs+交换+回溯，可以当模板套用：

class Solution {
    private List<List<Integer>> list = new ArrayList<>();        
    public List<List<Integer>> permute(int[] nums) {
        dfs(nums, 0, nums.length-1);
        return list;
    }
    public void dfs(int[] array,int start,int end) {
        if(start==end) {
            List<Integer> li = new ArrayList<>();
            for (int i = 0; i < array.length; i++) {
                li.add(array[i]);
            }
            list.add(li);
        } else {
            for (int i = start; i <= end; i++) {
                swap(array,start,i);
                dfs(array,start+1,end);
                swap(array,start,i);
            }
        }
    }
    public void swap(int[] array,int i,int j) {
        int temp = array[i];
        array[i] = array[j];
        array[j] = temp;
    }
}

142.环形链表 II

当快慢指针相遇之后，使用第三个指针从头开始和慢指针一起向后移动（步伐都是一步），当第三个指针与慢指针相遇时，即为环的入口：

public ListNode detectCycle(ListNode head) {
    ListNode fast = head;
    ListNode slow = head;
    ListNode res = head;
    while (fast != null && slow != null) {
        fast = fast.next;
        if (fast == null) {
            break;
        }
        fast = fast.next;
        slow = slow.next;
        if (slow == fast) {
            while (slow != res) {
                slow = slow.next;
                res = res.next;
            }
            return res;
        }
    }
    return null;
}

23.合并K个排序链表

结合前面那道合并两个链表，我们只需循环调用该方法即可：

class Solution {
    public ListNode mergeKLists(ListNode[] lists) {
        if (lists.length == 0) {
            return null;
        }
        if (lists.length == 1) {
            return lists[0];
        }
        ListNode res = lists[0];
        for (int i = 0; i < lists.length-1; i++) {
            res = mergeTwoLists(res, lists[i+1]);
        }
        return res;
    }
    public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
        if (l1 == null) {
            return l2;
        }
        if (l2 == null) {
            return l1;
        }
        if (l1.val < l2.val) {
            l1.next = mergeTwoLists(l1.next, l2);
            return l1;
        }else {
            l2.next = mergeTwoLists(l2.next, l1);
            return l2;
        }
    }
}

54.螺旋矩阵

模拟打印顺序：

public List<Integer> spiralOrder(int[][] matrix) {
    List<Integer> list = new ArrayList<>();
    int left = 0 ,up = 0;
    int right = matrix[0].length - 1;
    int down = matrix.length - 1;
    int size = matrix.length * matrix[0].length;
    while (list.size() < size) {
        for (int i = left; i <= right; i++) {
            if (list.size() < size) list.add(matrix[up][i]);
        }
        up++;
        for (int i = up; i <= down; i++) {
            if (list.size() < size) list.add(matrix[i][right]);
        }
        right--;
        for (int i = right; i >= left; i--) {
            if (list.size() < size) list.add(matrix[down][i]);
        }
        down--;
        for (int i = down; i >= up; i--) {
            if (list.size() < size) list.add(matrix[i][left]);
        }
        left++;
    }
    return list;
}

704.二分查找

如题，就是考二分

public int search(int[] nums, int target) {
    int l = 0, r = nums.length-1, m;
    while (l <= r) {
        m = (r + l)/2;
        if (target == nums[m]) {
            return m;
        }
        if (target > nums[m]) {
            l = m + 1;
        }else {
            r = m - 1;
        }
    }
    return -1;
}

232.用栈实现队列

两个栈实现，需要弹出时，将栈s1全部倒入s2，再从s2弹

class MyQueue {
    private Stack<Integer> s1 = new Stack<>();
    private Stack<Integer> s2 = new Stack<>();
    private int front;
    public MyQueue() {}
    public void push(int x) {
        if (s1.empty()) {
            front = x;
        }
        s1.push(x);
    }

    public int pop() {
        if (s2.empty()) {
            while(!s1.empty()) {
                s2.push(s1.pop());
            }
        }
        return s2.pop();
    }

    public int peek() {
        if (s2.empty()) {
            return front;
        }else {
            return s2.peek();
        }
    }

    public boolean empty() {
        return s1.empty() && s2.empty();
    }
}

42.接雨水

动规：从左向右，记录最大高度；再从右向左记录最大高度；最后每格的积水为两个数组中较小的那个再减去这个位置的高度：

public int trap(int[] height) {
    int res = 0;
    int len = height.length;
    int[] left = new int[len];
    int[] right = new int[len];
    left[0] = height[0];
    for (int i = 1; i < len; i++) {
        left[i] = Math.max(left[i-1], height[i]);
    }
    right[len-1] = height[len-1];
    for (int i = len - 2; i >= 0; i--) {
        right[i] = Math.max(height[i], right[i+1]);
    }
    for (int i = 0; i < len; i++) {
        res += Math.min(left[i], right[i]) - height[i];
    }
    return res;
}

94.二叉树的中序遍历

递归

class Solution {
    List<Integer> list = new ArrayList<>();
    public List<Integer> inorderTraversal(TreeNode root) {
        dfs(root);
        return list;
    }

    public void dfs (TreeNode root) {
        if (root != null) {
            dfs(root.left);
            list.add(root.val);
            dfs(root.right);
        }
    }
}

300.最长上升子序列

经典动规：（现在看看差不多又忘记怎么做了）

public int lengthOfLIS(int[] nums) {
    if (nums.length == 0) {
        return 0;
    }
    int[] dp = new int[nums.length];
    dp[0] = 1;
    int res = 1;
    for (int i = 1; i < nums.length; i++) {
        dp[i] = 1;
        for (int j = 0; j < i; j++) {
            if (nums[i] > nums[j]) {
                dp[i] = Math.max(dp[i], dp[j] + 1);
            }
        }
        res = Math.max(res, dp[i]);
    }
    return res;
}

143.重排链表

分三步：找中点——>翻转链表——>合并链表

class Solution {
    public void reorderList(ListNode head) {
        if (head == null) {
            return;
        }
        ListNode mid = middleList(head);
        ListNode l2 = reverseList(mid);
        mergeList(head, l2);
    }

    public ListNode middleList(ListNode head) {
        ListNode slow = head;
        ListNode fast = head;
        while (fast.next != null && fast.next.next != null) {
            fast = fast.next.next;
            slow = slow.next;
        }
        return slow;
    }

    public ListNode reverseList(ListNode head) {
        ListNode prev = null, next;
        while (head != null) {
            next = head.next;
            head.next = prev;
            prev = head;
            head = next;
        }
        return prev;
    }

    public void mergeList(ListNode l1, ListNode l2) {
        ListNode l1_t, l2_t;
        while (l1 != null && l2 != null) {
            l1_t = l1.next;
            l1.next = l2;
            l1 = l1_t;

            l2_t = l2.next;
            l2.next = l1;
            l2 = l2_t;
        }
    }
}

199.二叉树的右视图

向右深搜，技巧点：树的深度==集合大小，所以记录每层的深度，当depth == res.size()时为第一当这一层的最右节点：

class Solution {
    List<Integer> res = new ArrayList<>();

    public List<Integer> rightSideView(TreeNode root) {
        dfs(root, 0);
        return res;
    }

    public void dfs (TreeNode root, int depth) {
        if (root != null) {
            if (depth == res.size()) {
                res.add(root.val);
            }
            dfs(root.right, depth+1);
            dfs(root.left, depth+1);
        }
    }
}

70.爬楼梯

就是斐波那契数列，滚动数组解决：

public int climbStairs(int n) {
    if (n == 1) {
        return 1;
    }
    int[] dp = new int[3];
    dp[0] = 1;
    dp[1] = 1;
    for (int i = 2; i <= n; i++) {
        dp[2] = dp[0] + dp[1];
        dp[0] = dp[1];
        dp[1] = dp[2]; 
    }
    return dp[2];
}

22.链表中倒数第k个节点

两个指针，一个从起点，另一个从正数第k个移动：

public ListNode getKthFromEnd(ListNode head, int k) {
    ListNode fast = head;
    ListNode slow = head;
    while (k-- > 0) {
        fast = fast.next;
    }
    while (fast != null) {
        fast = fast.next;
        slow = slow.next;
    }
    return slow;
}

124.二叉树中的最大路径和

思路是递归，递归去找左右最大的路径和，个人觉得比较难，需重复做：

class Solution {
    int res = Integer.MIN_VALUE;
    public int maxPathSum(TreeNode root) {
        dfs(root);
        return res;
    }

    public int dfs(TreeNode node) {
        if (node == null) {
            return 0;
        }
        int left = Math.max(dfs(node.left), 0);
        int right = Math.max(dfs(node.right), 0);
        res = Math.max(res, node.val + left + right);
        return node.val + Math.max(left, right);
    }
}

69.x 的平方根

二分：

public int mySqrt(int x) {
    int res = -1, l = 0, r = x,m;
    while (l <= r) {
        m = l + (r - l)/2;
        if ((long)m*m <= x) {
            l = m + 1;
            res = m;
        }else {
            r = m - 1;
        }
    }
    return res;
}

56.合并区间

根据左端点排列后开始有序合并：

public int[][] merge(int[][] intervals) {
    List<int[]> res = new ArrayList<>();
    if (intervals == null || intervals.length == 0) {
        return res.toArray(new int[0][]);
    }
    if (intervals.length == 1) {
        return intervals;
    }
    Arrays.sort(intervals, (a, b) -> a[0] - b[0]);
    for (int i = 0; i < intervals.length-1;i++) {
        int left = intervals[i][0];
        int right = intervals[i][1];
        while (right >= intervals[i+1][1] || right >= intervals[i+1][0]) {
            if (right < intervals[i+1][1]) {
                right = intervals[i+1][1];
            }
            i++;
            if (i == intervals.length - 1) {
                break;
            }
        }
        int[] temp = new int[]{left, right};
        res.add(temp);
    }
    if (res.get(res.size() - 1)[1] < intervals[intervals.length -1][1]) {
        res.add(intervals[intervals.length -1]);
    }
    return res.toArray(new int[0][]);
}

19.删除链表的倒数第 N 个结点

我的写法可能麻烦些，主要是一些边界值的判断，思路和前面寻找倒数第N个节点思路一样，只不过是找N+1个，再做删除操作即可：

class Solution {
    public ListNode removeNthFromEnd(ListNode head, int n) {
            ListNode prev = findprev(head, n+1);
            if (prev == null) {
                if (head.next != null) {
                    return head.next;
                }
                return null;
            }
            if (prev.next != null) {
                prev.next = prev.next.next;
            }
            return head;
        }

    public ListNode findprev (ListNode head, int k) {
        ListNode fast = head;
        ListNode slow = head;
        while (k-- > 0) {
            if (fast == null) {
                return null;
            }
            fast = fast.next;
        }
        while (fast != null) {
            fast = fast.next;
            slow = slow.next;
        }
        return slow;
    }
}
最后修改：2021 年 11 月 20 日
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