leetcode 240: Search a 2D Matrix II C++Version

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

.
思路：
从右上角开始比较矩阵元素和target

1.如果target小于当前矩阵元素，则列减1，因为该元素是其所在列的最小元素，所以此列的剩余元素肯定不包含target

2.如果target大于当前矩阵元素，则行加1，因为该元素是其所在行的最大元素，所以此行的剩余元素肯定不包含target

Solution:

class Solution {

public:

    bool searchMatrix(vector<vector<int>>& matrix, int target) {

        int m=matrix.size(),n=matrix[0].size();

        int i=0,j=n-1;

        while(i < m && j >= 0){

            int x=matrix[i][j];

            if(x < target)

                ++i;

            else if( x > target)

                --j;

            else

                return true;

        }

        return false;

    }

};