LeetCode 42. Trapping Rain Water--c++解法

LeetCode 42. Trapping Rain Water–c++解法

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LeetCode题解专栏：LeetCode题解
我做的所有的LeetCode的题目都放在这个专栏里，大部分题目Java和Python的解法都有。

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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.

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!

Example:

Input: [0,1,0,2,1,0,1,3,2,1,2,1]
Output: 6

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这道题目跟第84题很类似，没做的可以先看这篇文章：LeetCode hard 84. Largest Rectangle in Histogram–python,java 15行,c++ 15行 解法

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c++ stack解法：

class Solution {
public:
int trap(vector<int>& height)
{
    int ans = 0, current = 0;
    stack<int> st;
    while (current < height.size()) {
        while (!st.empty() && height[current] > height[st.top()]) {
            int top = st.top();
            st.pop();
            if (st.empty())
                break;
            int distance = current - st.top() - 1;
            int bounded_height = min(height[current], height[st.top()]) - height[top];
            ans += distance * bounded_height;
        }
        st.push(current++);
    }
    return ans;
}
};

c++两边往中间夹的解法：

class Solution {
public:
int trap(vector<int>& height)
{
    int left = 0, right = height.size() - 1;
    int ans = 0;
    int left_max = 0, right_max = 0;
    while (left < right) {
        if (height[left] < height[right]) {
            height[left] >= left_max ? (left_max = height[left]) : ans += (left_max - height[left]);
            ++left;
        }
        else {
            height[right] >= right_max ? (right_max = height[right]) : ans += (right_max - height[right]);
            --right;
        }
    }
    return ans;
}
};