LeetCode: Trapping Rain Water

浏览数：27 / 时间：2015年06月11日

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.

技术分享

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!

菜鸟果然不会，看了别人的思路。

两个比较好的思路，一个是找出最高峰，然后分别从两边靠拢。在靠拢中记录当前的最高的，具体见http://www.cnblogs.com/lichen782/p/Leetcode_Trapping_Rain_Water.html

首先，找到最高的一个柱子，例如例子中的3。

技术分享

然后遍历左半边。从i=0开始靠近mid，遍历过程中维护一个curHeight，表示这当前遇到的最大的height值；如果当前的bar的height小于最大值，那么当前的bar能贡献的water就是height - curMaxHeight。否则的话，更新curMaxHeight。

为了更好理解这个算法，我们来track下这个过程：

技术分享

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;
    }
};

第二种思路是使用单调栈。这里是单调递减栈。

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;
    }