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74.cpp
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74.cpp
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// Author : Accagain
// Date : 17/4/10
// Email : [email protected]
/***************************************************************************************
*
* 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.
*
* 做法:
*
* 时间复杂度:
*
*
****************************************************************************************/
#include <cstdlib>
#include <cstdio>
#include <iostream>
#include <vector>
#include <string>
#define INF 0x3fffffff
using namespace std;
class Solution {
public:
int find(vector<vector<int>> matrix, int row_col, int target)
{
int l=0, mid, r, m=matrix.size(), n=matrix[0].size();
if(row_col == -1)
r = m-1;
else
r = n-1;
while(l<=r)
{
mid = (l+r)/2;
if(row_col == -1)
{
if(matrix[mid][0] > target)
r = mid -1;
else if(matrix[mid][n-1] < target)
l = mid + 1;
else
return mid;
}
else
{
if(matrix[row_col][mid] > target)
r = mid - 1;
else if(matrix[row_col][mid] < target)
l = mid + 1;
else
return mid;
}
}
return -1;
}
bool searchMatrix(vector<vector<int>>& matrix, int target) {
if(matrix.empty() || matrix[0].empty())
return false;
int row = find(matrix, -1, target);
if(row == -1)
return false;
int col = find(matrix, row, target);
if(col == -1)
return false;
return true;
}
};
int main() {
Solution *test = new Solution();
int data[] = {};
vector<int> x(data, data + sizeof(data) / sizeof(data[0]));
return 0;
}
//
// Created by cms on 17/4/10.
//