[leetcode] 74. Search a 2D Matrix

最新推荐文章于 2025-09-10 22:03:14 发布

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本文介绍了一种高效的搜索算法，用于在一特殊性质的二维矩阵中查找特定值。该矩阵的每一行从左到右递增排序，并且每行的第一个元素大于前一行的最后一个元素。文章提供了一个具体的例子和C++实现代码。

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.

解法一：

还是用二分法来做，用到的就是讲一个数值转化成矩阵元素的下表。另外要注意一些corner case，target小于matrix中第一个值或者大于matrix中最后一个值。

class Solution {
public:
    bool searchMatrix(vector<vector<int>>& matrix, int target) {
        int rows = matrix.size(), cols = matrix[0].size()?matrix[0].size():0;
        if(!rows || !cols) return 0;
        if(target<matrix[0][0]) return false;
        if(target>matrix.back().back()) return false;
        int left = 0, right = rows*cols - 1;
        
        while(left<=right){
            int mid = left + (right-left)/2;
            int i_mid = mid/cols, j_mid = mid%cols;
            if(matrix[i_mid][j_mid]==target) return true;
            else if(matrix[i_mid][j_mid]>target) right = mid-1;
            else left = mid+1;
        }

        return false;
        
    }
};