标签：
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!
思路:
这一道题类似于leetcode 11 题Container With Most Water. 要求最多能装多少水.有两种思路.
思路一:
先从左往右遍历.每遍历一个数字,就贪心得到这个数字所在的格子能最多装水. 由于是从左到右遍历,我们只考虑的左边的高度,没有考虑右边的高度(实际应该选择两者小的那个). 因此需要再从右往左遍历,同样贪心这个数字所在格子
最多能装多少水,这次以右边高度为准. 最终综合从左到右和从右到左的结果,取两者小的.
代码:(runtime 19ms)
1 class Solution { 2 public: 3 int trap(vector<int>& height) { 4 vector<int> leftToRight; 5 vector<int> rightToLeft; 6 aux_function(height.begin(), height.end(), leftToRight); 7 8 aux_function(height.rbegin(), height.rend(), rightToLeft); 9 10 int ans = 0; 11 auto it1 = leftToRight.begin(); 12 auto it2 = rightToLeft.rbegin(); 13 14 while (it1 != leftToRight.end()){ 15 ans += min(*it1++, *it2++); 16 } 17 return ans; 18 } 19 20 private: 21 template<class Iter> 22 void aux_function(Iter begin, Iter end, vector<int> &ret) { 23 int left = 0; 24 for (auto it = begin; it != end; it++) { 25 left = left > *it ? left : *it; 26 ret.push_back(left - *it); 27 } 28 } 29 };
思路二:
如上图中所示, 先求黑色格子与蓝色格子的总面积,然后再减去黑色各自面积.
代码:(runtime 10ms)
class Solution { public: int trap(vector<int> A) { int n = A.size(); int summap = 0; int sumtot = 0; for(int i = 0; i < n; i++) summap += A[i]; int left = 0, right = n - 1; int leftbar = 0, rightbar = 0; while(left <= right) { leftbar = max(A[left], leftbar); rightbar = max(A[right], rightbar); if(leftbar <= rightbar) { sumtot += leftbar; left++; //right--; } else { sumtot += rightbar; right--; //left++; } } return sumtot - summap; } };
leetcode problem 42 -- Trapping Rain Water
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原文地址：http://www.cnblogs.com/lysuns/p/4481918.html
