接雨水

問題描述：

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.
如上圖所示，海拔分別爲 [0,1,0,2,1,0,1,3,2,1,2,1], 返回 6.

挑戰

O(n) 時間, O(1) 空間

思路：題目不難，觀察下就可以發現被水填滿後的形狀是先升後降的塔形，因此，先遍歷一遍找到塔頂，然後分別從兩邊開始，往塔頂所在位置遍歷，水位只會增高不會減小，且一直和最近遇到的最大高度持平，這樣知道了實時水位，就可以邊遍歷邊計算面積。
c++代碼如下：

class Solution {
public:
    /*
     * @param heights: a list of integers
     * @return: a integer
     */
    int trapRainWater(vector<int> &heights) {
        // write your code here
        //防止越界
        if(heights.size() < 2)return 0;
        int maxheight = 0;
        int maxindex = 0;
        //尋找最高峯
        for(int i = 0;i!=heights.size();++i){
            if(heights[i]>maxheight){
                maxheight = heights[i];
                maxindex = i;
            }
        }
        int waterarea = 0;
        int pedge = 0;
        int qedge = 0;
        for(;qedge<maxindex;){
            qedge++;
            if(heights[qedge] > heights[pedge]){
                pedge = qedge;
            }
            else{
                waterarea = waterarea+heights[pedge]-heights[qedge];
            }
        }
        pedge = heights.size() - 1;
        qedge = heights.size() - 1;
        for(;qedge>maxindex;){
            qedge--;
            if(heights[qedge] > heights[pedge]){
                pedge = qedge;
            }
            else{
                waterarea = waterarea+heights[pedge]-heights[qedge];
            }
        }
        return waterarea;
    }
};