Leetcode74. Search 2D Matrix

题目传送地址： https://leetcode.cn/problems/search-a-2d-matrix/

运行效率

代码如下：

class Solution {
    
  public static boolean searchMatrix(int[][] matrix, int target) {
    
        int row;  //target所在的行
        //处理边界情况
        if (target > matrix[matrix.length - 1][0]) {
    
            if (target > matrix[matrix.length - 1][matrix[0].length - 1]) {
    
                return false;
            } else {
    
                row = matrix.length - 1;
            }
        }
        //First use the dichotomy method to determine the target value possible line
        int left = 0;
        int right = matrix.length;
        while (left != right) {
    
            int mid = (left + right) / 2;
            if (target == matrix[mid][0]) {
    
                return true;
            }
            if (target > matrix[mid][0]) {
    
                if(left==mid){
      //如果left指针和rightWhen the pointer is encountered, it can jump out of the loop
                    break;
                }
                left = mid;
            }
            if (target < matrix[mid][0]) {
    
                if(right==mid){
     //如果left指针和rightWhen the pointer is encountered, it can jump out of the loop
                    break;
                }
                right = mid;
            }
        }
        row = left;

        //Then use the dichotomy method to determine the target value possible columns
        left = 0;
        right = matrix[0].length;
        while (left != right) {
    
            int mid = (left + right) / 2;
            if (target == matrix[row][mid]) {
    
                return true;
            }
            if (target > matrix[row][mid]) {
    
                if(left==mid){
     //如果left指针和rightWhen the pointer is encountered, it can jump out of the loop
                    break;
                }
                left = mid;
            }
            if (target < matrix[row][mid]) {
    
                if(right==mid){
     //如果left指针和rightWhen the pointer is encountered, it can jump out of the loop
                    break;
                }
                right = mid;
            }
        }
        return false;
    }
}