Search a 2D Matrix

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.

For example,

Consider the following matrix:

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

Given target = 3, return true.

C++实现代码如下：

#include<iostream>
#include<vector>
using namespace std;

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

int main()
{
    Solution s;
    vector<vector<int> > vec=
    {
        {-10,-9},
        {-7,-6},
        {-5,-4},
        {-3,-2}
    };
    cout<<s.searchMatrix(vec,-6)<<endl;
}

开始提交了一种，死活通不过。

class Solution {
public:
    bool searchMatrix(vector<vector<int> > &matrix, int target) {
        if(matrix.empty()||matrix[0].empty())
            return false;
        size_t i=0;
        size_t j=matrix[0].size()-1;
        int temp;
        while(i<matrix.size()&&j>=0)
        {
            temp=matrix[i][j];
            if(target==temp)
                return true;
            if(target<temp)
            {
                j--;
                continue;
            }
            if(target>temp)
            {
                i++;
                continue;
            }
        }
        if(i>=matrix.size()||j<0)
            return false;
        return true;
    }
};

一直报错Last executed input:[[-10],[-7],[-4]], -6

就因为将i和j声明为size_t类型，可能出现下溢。可以参考：https://oj.leetcode.com/discuss/11366/why-is-the-last-executed-input-error