[数据结构与算法]算法打卡，用于自律

题目一

?解法

/**
 * 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 {
    public int countNodes(TreeNode root) {
        TreeNode l = root, r = root;
        int hl = 0, hr = 0;
        while (l != null) {
            l = l.left;
            hl++;
        }
        while (r != null) {
            r = r.left;
            hl++;
        }
        if (hl == hr) {
            return (int)Math.pow(2, hl) - 1;
        }
        return 1+countNodes(root.left)+countNodes(root.right);
    }
}

题目二

解法

class Solution {
    public void rotate(int[][] matrix) {
        int n = matrix.length;
        for (int i = 0; i < n; i++) {
            for (int j = i; j < n; j++) {
                int temp = matrix[i][j];
                matrix[i][j] = matrix[j][i];
                matrix[j][i] = temp;
            }
        }
        for(int[] q : matrix){
            for(int i = 0;i<q.length/2;i++){
                int temp = q[i];
                q[i] = q[q.length-i-1];
                q[q.length-i-1] = temp;
            }
        }
    }
}

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