leetcode 74. Search a 2D Matrix

阳洞洞2025-05-20 9:20

题目描述

要求时间复杂度必须是log(m*n)。那么对每一行分别执行二分查找就不符合要求，这种做法的时间复杂度是m*log(n)。

方法一，对每一行分别执行二分查找：

cpp 复制代码

class Solution {
public:
    bool searchMatrix(vector<vector<int>>& matrix, int target) {
        int m = matrix.size();
        int n = matrix[0].size();
        for(int i = 0;i < m;i++){
            if(binary_search(matrix[i],0,n-1,target))
                return true;
        }
        return false;
    }

    bool binary_search(vector<int> row,int left,int right,int target){
        int mid = 0;
        while(left <= right){
            mid = left + ((right-left)>1);
            if(row[mid] == target)
                return true;
            else if(row[mid] > target){
                right = mid-1;
            }else{
                left = mid+1;
            }
        }
        return false;
    }
};

方法二，对整个矩阵执行二分查找，关键是要将整体的序号映射到行和列的下标：

时间复杂度log(m*n)，符合要求。

cpp 复制代码

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