[LeetCode] Trapping Rain Water 解题报告

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!

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[解题思路]
对于任何一个坐标，检查其左右的最大坐标，然后相减就是容积。所以，
1. 从左往右扫描一遍，对于每一个坐标，求取左边最大值。
2. 从右往左扫描一遍，对于每一个坐标，求最大右值。
3. 再扫描一遍，求取容积并加和。
#2和#3可以合并成一个循环，


[Code]

1:    int trap(int A[], int n) {  
2:      // Start typing your C/C++ solution below  
3:      // DO NOT write int main() function  
4:      if(n<2) return 0;  
5:      int *maxL = new int[n], *maxR=new int[n];  
6:      int max = A[0];  
7:      maxL[0] =0;  
8:      for(int i =1; i<n-1; i++)  
9:      {  
10:        maxL[i] =max;  
11:        if(max < A[i])  
12:          max = A[i];        
13:      }  
14:      max=A[n-1];  
15:      maxR[n-1]=0;  
16:      int ctrap,ttrap=0;  
17:      for(int i = n-2; i>0; i--)  
18:      {  
19:        maxR[i] = max;  
20:        ctrap = min(maxL[i], maxR[i]) -A[i];  
21:        if(ctrap>0)  
22:          ttrap+=ctrap;  
23:        if(max<A[i])  
24:          max = A[i];  
25:      }  
26:      delete maxL, maxR;  
27:      return ttrap;  
28:    }  

如果我是面试官的话，在这里可以加一个变形。不求所有容积，而是求容积中的最大值。这样就类似于Container With Most Water，但是又有不同的地方。这题里面每一个坐标本身是占体积的。所以从两边往中间扫的时候，根本不知道中间坐标共占用了多少体积。