[LeetCode]240. Search a 2D Matrix II

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本文介绍了一种高效的矩阵搜索算法，该算法能在排序矩阵中快速查找指定数值。矩阵的每一行及每一列都按升序排列。文章通过示例矩阵展示了如何逐行遍历并使用二分查找来确定目标值是否存在。

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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 in ascending from left to right.
Integers in each column are sorted in ascending from top to bottom.

For example,

Consider the following matrix:

[
  [1,   4,  7, 11, 15],
  [2,   5,  8, 12, 19],
  [3,   6,  9, 16, 22],
  [10, 13, 14, 17, 24],
  [18, 21, 23, 26, 30]
]

Given target = 5, return true.

Given target = 20, return false.

思路：每行进行遍历，判断每行第一个是否小于目标，如果是则对该行进行二分查找，如果不是则说明这行以及下边所有行都不会存在目标，返回false

public class Solution {
    public boolean searchMatrix(int[][] matrix, int target) {
        if(matrix.length==0){
            return false;
        }
        if(matrix[0].length==0){
            return false;
        }
        for(int i=0;i<matrix.length;i++){
            if(target<matrix[i][0]){
                return false;
            }
            if(binarySearch(matrix[i],0,matrix[i].length-1,target)){
                return true;
            }
        }
        return false;
    }
    
    public boolean binarySearch(int[] nums,int start,int end,int target){
        if(start>end){
            return false;
        }
        int medium=(start+end)/2;
        if(target==nums[start]||target==nums[end]||target==nums[medium]){
            return true;
        }else{
            if(target>nums[medium]){
                return binarySearch(nums,medium,end-1,target);
            }else{
                return binarySearch(nums,start+1,medium,target);
            }
            
        }
    }
}