[240]Search a 2D Matrix II

【题目描述】

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】

如果从第一个元素开始实现比较麻烦，转换思路，从右上角的元素开始，发现不管哪个元素其左边的元素都比它小，下边的元素都比它大，如果target大于该元素，则搜索其下边的元素，如果小于则搜索其左边的元素。

【代码】

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