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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
Search a 2D Matrix
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