[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
Given
[0,1,0,2,1,0,1,3,2,1,2,1]
, return 6
.![](http://www.leetcode.com/wp-content/uploads/2012/08/rainwatertrap.png)
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!
» Solve this problem[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。