代码随想录算法训练营第三一天| 56. 合并区间 738.单调递增的数字 968.监控二叉树

今日任务

56. 合并区间  
738.单调递增的数字 
968.监控二叉树

56. 合并区间 

题目链接： . - 力扣（LeetCode）

class Solution {
    public int[][] merge(int[][] intervals) {
        Arrays.sort(intervals, (a,b)->a[0] - b[0]);
        if (intervals.length == 1) {
            return intervals;
        }

        int left = intervals[0][0];
        int right = intervals[0][1];
        List<int[]> res = new ArrayList<>();
        for (int i = 1; i < intervals.length; i++) {
            if (intervals[i][0] > right) {
                res.add(new int[]{left, right});
                left = intervals[i][0];
                right = intervals[i][1];
            } else {
                right = Math.max(right, intervals[i][1]);
            }
        }
        res.add(new int[]{left, right});
        int[][] result = new int[res.size()][2];
        for (int i = 0; i < res.size(); i++) {
            result[i] = res.get(i);
        }
        return result;
    }
}

738.单调递增的数字 

题目链接： . - 力扣（LeetCode）

class Solution {
    public int monotoneIncreasingDigits(int n) {
        String s = String.valueOf(n);
        char[] charArray = s.toCharArray();
        int start = charArray.length;
        for (int i = s.length() - 2; i >= 0; i--) {
            if (charArray[i] > charArray[i + 1]) {
                start = i + 1;
                charArray[i]--;
            }
        }
        for (int i = start; i < charArray.length; i++) {
            charArray[i] = '9';
        }
        return Integer.parseInt(String.valueOf(charArray));
    }
}

968.监控二叉树

题目链接： . - 力扣（LeetCode）

/**
 * Definition for a binary tree node.
 * public 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;
 *     }
 * }
 */
class Solution {
    int res = 0;
    public int minCameraCover(TreeNode root) {
        if (minCamera(root) == 0) {res++;}
        return res;
    }

    public int minCamera(TreeNode root) {
        /*
        * 0 表示该节点无覆盖
        * 1 表示该节点有摄像头
        * 2 表示该节点无摄像头，但有覆盖
        * */
        if (root == null) {return 2;}
        int left = minCamera(root.left);
        int right = minCamera(root.right);

        if (left == 2 && right == 2) { return 0;}
        else if (left == 0 || right == 0) {res++; return 1;}
        else {return 2;}
    }
}