Search a 2D Matrix

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题目要求：

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.

示例：

[  [1,   3,  5,  7],  [10, 11, 16, 20],  [23, 30, 34, 50]]

Given target = 3, return true.

代码实现：

#include<stdafx.h>#include <iostream>#include<vector>using namespace std;class Solution {public:bool searchMatrix(vector<vector<int> > &matrix, int target) {        size_t m=matrix.size();size_t n=matrix[0].size();size_t i=0;size_t j=n-1;//从右上角开始找，如果target大于此数，二维矩阵往下找，否则往左找while(j>=0&&i<m){if(matrix[i][j]<target){i++;}else if (matrix[i][j]>target){if(j==0)return false;//此处防止j=0时，j--变成4294967295j--;}elsereturn true;}return false;      }};void main(){           //初始化二维矩阵    vector< vector<int>> matrix(3,vector<int>(4,0)); matrix[0][0]=1;matrix[0][1]=3;matrix[0][2]=5;matrix[0][3]=7;matrix[1][0]=10;matrix[1][1]=11;matrix[1][2]=16;matrix[1][3]=20;matrix[2][0]=23;matrix[2][1]=30;matrix[2][2]=34;matrix[2][3]=50;//第一次验证没有通过的一组              /* vector< vector<int>> matrix(3,vector<int>(1,0)); matrix[0][0]=-11;matrix[1][0]=-7;matrix[2][0]=-5;*/Solution s;cout<<s.searchMatrix(matrix,-6)<<endl;getchar();}

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