Search a 2D Matrix----easy-优快云博客

本文介绍了一种在具有特定排序特性的二维矩阵中搜索指定值的高效算法。通过遍历矩阵行并利用二分查找定位目标值，该算法能够快速确定目标值是否存在。讨论了在目标值位于矩阵区间内时的二分查找优化，特别注意了循环结束条件的正确实现，避免了在边界情况下的逻辑错误。

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Write an efficient algorithm that searches for a value in an m x n matrix. This matrix has the following properties:

Integers in each row are sorted from left to right.
The first integer of each row is greater than the last integer of the previous row.

For example,

Consider the following matrix:

[
  [1,   3,  5,  7],
  [10, 11, 16, 20],
  [23, 30, 34, 50]
]

Given target = 3, return true.

题目不难代码如下

public class Sa2M {
	public static void main(String[] args){
		//int matrix[][] = {{1,3,5,7},{10,11,16,20},{23,30,34,50}};
		int matrix[][] = {{1,3}};
		int target = 2;
		System.out.println(searchMatrix(matrix,target));
	}
	
	public static boolean searchMatrix(int[][] matrix, int target) {
		int m,n;
		m = matrix.length;
		n = matrix[0].length;
		int mid;
		int low,high;
		for(int i=0;i<m;i++){
			if(target>= matrix[i][0] && target<= matrix[i][n-1]){
				low = 0;
				high = n-1;
				while(low<=high){
					mid = (low +high)/2;
					if(matrix[i][mid] < target){
						low = mid+1;
					}else if(matrix[i][mid] >target){
						high = mid-1;
					}
					else{
						return true;
					}
				}
			}
		}
		return false;
    }
}

遍历二维数组行，找到目标值所在区间，然后二分查找

太久没做题什么都慢，二分的时候low和high的更新中直接赋值为mid，结果发现在low和high相差为1的时候循环无法结束了

更新的时候mid值已经做过判断，新区间内不应包含mid值，同时在low和high相差1时再次进入循环两值相等，再再次就无法进入循环退出了