42. 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.

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


贪心题目
计算水有多少,就是找高低落差;
取左右指针,往中心靠近,每次找值最小的靠近;
用level来计算目前遇到过的最高的bar,这样后面遇到较低的bar,相减就是能容纳的水

class Solution {
public:
    int trap(vector<int>& height) {
        int l=0,r=height.size()-1,level=0,water=0;
        while(l<r)
        {
            int lower=height[height[l]<height[r]?l++:r--];
            level=max(level,lower);
            water+=level-lower;
        }
        return water;
    }
};

42. Trapping Rain Water

原文：https://www.cnblogs.com/lychnis/p/11705032.html