74.Search a 2D Matrix

最新推荐文章于 2025-05-19 23:44:29 发布

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最新推荐文章于 2025-05-19 23:44:29 发布

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分类专栏： leetcode

本文链接：https://blog.youkuaiyun.com/qq_27655147/article/details/79356204

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本文介绍了一种高效的矩阵搜索算法，该算法适用于行和列都按升序排列的矩阵，通过两次二分查找快速定位目标值。示例中给出了具体的实现代码。

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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.

思路：用两次二分查找，先对行查找再对列查找

class Solution {
public:
    bool searchMatrix(vector<vector<int>>& matrix, int target) {
        int left = 0;  
        int right = matrix.size()-1;  
        if(left != right)  
        {  
            while(left<=right)  
            {  
                int middle = left + (right-left)/2;  
                if(matrix[middle][0] < target)  
                {  
                    left = middle+1;  
                }  
                else if(matrix[middle][0] > target)  
                {  
                    right = middle-1;  
                }  
                else  
                {  
                    return true;  
                }  
            }  
        }  
        if(right == -1)  
        {  
            return false;  
        }  
        else  
        {  
            int row = right;  
            int left = 0;  
            int right = matrix[row].size()-1;  
            while(left<=right)  
            {  
                int middle = left + (right-left)/2;  
                if(matrix[row][middle] < target)  
                {  
                    left = middle+1;  
                }  
                else if(matrix[row][middle] > target)  
                {  
                    right = middle-1;  
                }  
                else  
                {  
                    return true;  
                }  
            }  
            return false;  
        }  
          
    }
};