leetcode 240 : Search a 2D Matrix II

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 in ascending from left to right.
Integers in each column are sorted in ascending from top to bottom.
For example,

Consider the following matrix:

[
  [1,   4,  7, 11, 15],
  [2,   5,  8, 12, 19],
  [3,   6,  9, 16, 22],
  [10, 13, 14, 17, 24],
  [18, 21, 23, 26, 30]
]

Given target = 5, return true.

Given target = 20, return false.

代码:

class Solution {
public:
    //由于行增和列增的性质，可从左下角开始，向上递减，向右递增
    //以左下为起点，往右上搜索，如果target比当前m[row][column]小，则row--,否则column++;相等返回true
    //时间复杂度O(m+n)
    bool searchMatrix(vector<vector<int>>& matrix, int target) {
        int row=matrix.size()-1,column=0;
        while(row>=0 && column<matrix[0].size()){
            if(target<matrix[row][column]){
                row--;
            }else if(target>matrix[row][column]){
                column++;;
            }else{
                return true;
            }
        }
        return false;
    }
};