Leetcode | Trapping Rain Water

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

这道题用的和Largest Rectangle in Histogram类似的算法，都是用了一个栈。如果比栈顶小或者栈为空，就push进栈；如果比栈顶大，就计算围起来的面积。为了不重复计算面积，每次只计算，在栈顶以上，在栈顶的前一个数(pop之后的栈顶）和当前数围起来的长方形面积(Line 13)。

class Solution {
public:
    int trap(int A[], int n) {
        int sum = 0; 
        stack<int> st;
        
        for (int i = 0; i < n; ) {
            if (st.empty()) {
                st.push(i++);
            } else if (A[i] >= A[st.top()]) {
                int tmp = st.top();
                st.pop();
                if (!st.empty()) sum += (i - st.top() - 1) * (min(A[i], A[st.top()]) - A[tmp]);
            } else {
                st.push(i++);
            }
        }
        return sum;
    }
};

Method II 

网上有另外一种算法，思路是算出每个位置可以存的雨量。就是把总雨量分摊到每个位置。当前位置的雨量由它的min(左边最大值,右边最大值)决定。照着思路重写了一遍。

class Solution {
public:
    int trap(int A[], int n) {
        if (n == 0) return 0;
        int sum = 0; 
        vector<int> lfmost(n, 0);
        
        lfmost[0] = A[0];
        for (int i = 1; i < n; ++i) {
            lfmost[i] = lfmost[i - 1] > A[i - 1] ? lfmost[i - 1] : A[i - 1];
        }
        
        int max = A[n - 1];
        for (int j = n - 2; j >= 1; --j) {
            if (A[j + 1] > max) max = A[j + 1];
            int s = min(lfmost[j], max) - A[j];
            if (s > 0) sum += s;
        }
        return sum;
    }
};