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.

分析

我们从第一行i，最后一列j开始搜索

观察到该搜索具有两个重要性质，若matrix[i][j] > target,说明第i行之下的j列的所有元素都大于target,target必在j列的左侧；

若matrix[i][j] < target,说明第j列前的第i行的所有元素都小于target，target必在i行的下面。

代码

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