【LeetCode】74. Search a 2D Matrix

最新推荐文章于 2025-05-19 23:44:29 发布

转载最新推荐文章于 2025-05-19 23:44:29 发布 · 53 阅读

本文介绍了一种高效的算法，用于在一个m x n的矩阵中搜索特定值。该矩阵每一行的整数从左到右排序，并且每行的第一个整数大于前一行的最后一个整数。通过纵向二分查找确定目标值所在的行，再进行常规的二分查找。

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Search a 2D Matrix

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：纵向二分查找确定target所在行，再进行普通的二分查找。

class Solution {
public:
    bool searchMatrix(vector<vector<int> > &matrix, int target) {
        if(matrix.empty() || matrix[0].empty())
            return false;
        int m = matrix.size();
        int n = matrix[0].size();
        
        int low = 0;
        int high = m-1;
        while(low <= high)
        {
            int mid = low+(high-low)/2;
            if(matrix[mid][0] == target)
                return true;
            else if(matrix[mid][0] > target)
                high = mid-1;
            else
                low = mid+1;
        }
        low --;
        if(low < 0)
            return false;
        int left = 0;
        int right = n-1;
        while(left <= right)
        {
            int mid = left+(right-left)/2;
            if(matrix[low][mid] == target)
                return true;
            else if(matrix[low][mid] > target)
                right = mid-1;
            else
                left = mid+1;
        }
        
        return false;
    }
};