Leetcode | Trapping Rain Water

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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!

Method I

這道題用的和Largest Rectangle in Histogram類似的演算法，都是用了一個棧。如果比棧頂小或者棧為空白，就push進棧；如果比棧頂大，就計算圍起來的面積。為了不重複計算面積，每次只計算，在棧頂以上，在棧頂的前一個數(pop之後的棧頂）和當前數圍起來的長方形面積(Line 13)。

 1 class Solution { 2 public: 3     int trap(int A[], int n) { 4         int sum = 0;  5         stack<int> st; 6          7         for (int i = 0; i < n; ) { 8             if (st.empty()) { 9                 st.push(i++);10             } else if (A[i] >= A[st.top()]) {11                 int tmp = st.top();12                 st.pop();13                 if (!st.empty()) sum += (i - st.top() - 1) * (min(A[i], A[st.top()]) - A[tmp]);14             } else {15                 st.push(i++);16             }17         }18         return sum;19     }20 };

Method II 

網上有另外一種演算法，思路是算出每個位置可以存的雨量。就是把總雨量分攤到每個位置。當前位置的雨量由它的min(左邊最大值,右邊最大值)決定。照著思路重寫了一遍。

 1 class Solution { 2 public: 3     int trap(int A[], int n) { 4         if (n == 0) return 0; 5         int sum = 0;  6         vector<int> lfmost(n, 0); 7          8         lfmost[0] = A[0]; 9         for (int i = 1; i < n; ++i) {10             lfmost[i] = lfmost[i - 1] > A[i - 1] ? lfmost[i - 1] : A[i - 1];11         }12         13         int max = A[n - 1];14         for (int j = n - 2; j >= 1; --j) {15             if (A[j + 1] > max) max = A[j + 1];16             int s = min(lfmost[j], max) - A[j];17             if (s > 0) sum += s;18         }19         return sum;20     }21 };