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42. Trapping Rain Water Problem's Link
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Mean:
在坐标上给你一些竖直放置的条形积木,问你这个积木能够容纳多少液体.
analyse:
首先找出最高的积木,然后从前往后一直扫到最高积木,从后往前一直扫到最高积木,两部分体积相加即可.
Time complexity: O(N)
view code
* -----------------------------------------------------------------
* Copyright (c) 2016 crazyacking.All rights reserved.
* -----------------------------------------------------------------
* Author: crazyacking
* Date : 2016-03-05-22.10
*/
#include <queue>
#include <cstdio>
#include <set>
#include <string>
#include <stack>
#include <cmath>
#include <climits>
#include <map>
#include <cstdlib>
#include <iostream>
#include <vector>
#include <algorithm>
#include <cstring>
using namespace std;
typedef long long( LL);
typedef unsigned long long( ULL);
const double eps( 1e-8);
class Solution
{
public :
int trap( vector < int >& height)
{
int len = height . size (), front = 0 , res = 0;
int max_height = 0 , max_idx = len - 1;
for( int i = 0; i < len; ++ i)
{
if( max_height < height [ i ])
{
max_height = height [ i ];
max_idx = i;
}
}
while( front < max_idx)
{
while( front + 1 < max_idx && height [ front ] < height [ front + 1 ])
++ front;
int stone = 0;
int back = front + 1;
while( back < max_idx && height [ back ] < height [ front ])
{
stone += height [ back ];
++ back;
}
res +=( back - front - 1) * height [ front ] - stone;
front = back;
}
int back = len - 1;
while( back > max_idx)
{
while( back - 1 > max_idx && height [ back ] < height [ back - 1 ])
-- back;
int stone = 0;
int front = back - 1;
while( front > max_idx && height [ front ] < height [ back ])
{
stone += height [ front ];
-- front;
}
res +=( back - front - 1) * height [ back ] - stone;
back = front;
}
return res;
}
};
int main()
{
freopen( "H: \\ Code_Fantasy \\ in.txt" , "r" , stdin);
int n;
while( cin >>n)
{
vector < int > ve;
for( int i = 0; i <n; ++ i)
{
int tempVal;
cin >> tempVal;
ve . push_back( tempVal);
}
Solution solution;
cout << solution . trap( ve) << endl;
}
return 0;
}
/*
*/
2.方法二:时间O(n),空间(1)
{
public :
int trap( vector < int >& height)
{
int left = 0 , right = height . size() - 1;
int maxLeft = 0 , maxRight = 0;
int res = 0;
while( left < right)
{
if( height [ left ] <= height [ right ])
{
if( height [ left ] > maxLeft)
maxLeft = height [ left ];
res += maxLeft - height [ left ];
left ++;
}
else
{
if( height [ right ] > maxRight)
maxRight = height [ right ];
res += maxRight - height [ right ];
right --;
}
}
return res;
}
};
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