题目

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.

分析

思路就是二分法，不同的是查找时需要将元素的序号映射为其在矩阵中的坐标。

实现

class Solution {
public:
    bool searchMatrix(vector<vector<int>>& matrix, int target) {
        m=matrix.size();
        n=matrix[0].size();
        return helper(matrix, target, 0, m*n-1);
    }
private:
    int m,n;
    bool helper(vector<vector<int>>& matrix, int target, int start, int end){
        int mid = (start + end) / 2;
        int x = mid / n;
        int y = mid % n;
        if(start>=end) return matrix[x][y]==target;
        if(matrix[x][y]<target)
            return helper(matrix, target, mid+1, end);
        else
            return helper(matrix, target, start, mid);
    }
};

思考

不适用递归会更快。