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42.cpp
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// Author : Accagain
// Date : 17/3/26
// Email : chenmaosen0@gmail.com
/***************************************************************************************
* Given n non-negative integers representing an elevation map where the width of each bar is 1,
*
* compute how much water it is able to trap after raining.
*
* For example,
*
* Given [0,1,0,2,1,0,1,3,2,1,2,1], return 6.
*
* The above elevation map is represented by array [0,1,0,2,1,0,1,3,2,1,2,1]. In this case, 6 units of rain water (blue section) are being trapped. Thanks Marcos for contributing this image!
*
* 做法:
* Max * O(n)的做法 超时;
* 左右扫,补短的,维护左右最大值;
*
*
* 时间复杂度:
* O(n^2)
*
****************************************************************************************/
#include <cstdlib>
#include <cstdio>
#include <iostream>
#include <vector>
#include <string>
#define INF 0x3fffffff
using namespace std;
class Solution {
public:
/*
int trap(vector<int>& height) {
int ans = 0;
for(int he=1; he<INF; he++)
{
int be = -1;
for(int i=0; i<height.size(); i++)
{
if(height[i]>=he)
{
if(be != -1)
ans += (i-be-1);
be = i;
}
}
if(be == -1)
break;
}
return ans;
}
*/
int trap(vector<int>& height)
{
int ans = 0;
int left_most = 0, right_most = 0, left = 0, right = height.size()-1;
while(left<=right)
{
if(height[left] <= height[right])
{
if(height[left] < left_most)
{
ans += left_most - height[left];
}
else
left_most = max(left_most, height[left]);
left ++;
}
else
{
if(height[right] < right_most)
{
ans += right_most - height[right];
}
else
right_most = max(right_most, height[right]);
right --;
}
}
return ans;
}
};
int main() {
Solution *test = new Solution();
int data[] = {0,1,0,2,1,0,1,3,2,1,2,1};
vector<int> x(data, data + sizeof(data) / sizeof(data[0]));
printf("%d\n", test->trap(x));
return 0;
}
//
// Created by cms on 17/3/26.
//