LeetCode: Trapping Rain Water

標籤：

Title:

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!

 

菜鳥果然不會，看了別人的思路。

兩個比較好的思路，一個是找出最高峰，然後分別從兩邊靠攏。在靠攏中記錄當前的最高的，具體見http://www.cnblogs.com/lichen782/p/Leetcode_Trapping_Rain_Water.html

首先，找到最高的一個柱子，例如例子中的3。

然後遍曆左半邊。從i=0開始靠近mid，遍曆過程中維護一個curHeight，表示這當前遇到的最大的height值；如果當前的bar的height小於最大值，那麼當前的bar能貢獻的water就是height - curMaxHeight。否則的話，更新curMaxHeight。

為了更好理解這個演算法，我們來track下這個過程：

     

         

 

 

class Solution{public:    int trap(vector<int> & height){        if (height.size() <=1 )            return 0;        int n = height.size();        int hi_index = 0;        for (int i = 0 ; i < height.size(); i++){            if (height[i]> height[hi_index]){                hi_index= i;            }        }        int cur_h = 0;        int water = 0;        for (int i = 0 ; i < hi_index; i++){            if (height[i] > cur_h)                cur_h = height[i];            else{                water += (cur_h - height[i]);            }        }        cur_h = 0;        for (int i = height.size()-1; i > hi_index; i--){            if (height[i] > cur_h)                cur_h = height[i];            else                water += (cur_h - height[i]);        }        return water;    }};

第二種思路是使用單調棧。這裡是單調遞減棧。

int trap(vector<int> & height) {        stack<int> stk; // descending        int result = 0;        int i = 0;        int n = height.size();        while (i < n) {            if (stk.size() == 0 || height[stk.top()] > height[i]) {                stk.push(i); // the index                i++;            } else { // A[i] >= stk.top();                int j = stk.top();                stk.pop();                if (stk.size() != 0) {                    result += (i - stk.top() - 1) * (min(height[stk.top()], height[i]) - height[j]);                }            }        }        return result;    }

 

LeetCode: Trapping Rain Water