[leetcode] 42. Trapping Rain Water 解题报告

题目链接: https://leetcode.com/problems/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!

思路: two point指向左右两端, 维护一个当前左右同时能达到的最大的高度, 并且让面积小的一端先移动, 如果遇到一个新的高度小于当前最高高度, 那么其高度差就可以存储一些水. 如果碰到一个高度, 使得左右能同时达到的高度增加, 则更新.

代码如下: 

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