[LeetCode]Trapping Rain Water

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：找到最大值及其位置pos。然后从左往右遍历到pos，再从右往左遍历到pos。

什么时候会加水呢？

遍历的时候记录最大值，也就是全局第二大的值，如果当前值比第二大的小，那么就会加水了，小多少就加多少。

加水是一列列加的。下图。

class Solution {
public:
    int trap(vector<int>& height) {
        if(height.size()<=2) return 0;
        int result=0;
        int pos=0,maxHeight=0;
        for(int i=0;i<height.size();i++)
        {
            if(height[i]>maxHeight)
            {
                maxHeight = height[i];
                pos = i;
            }
        }
        int cur_max = height[0];
        for(int i=1;i<pos;i++)
        {
            cur_max = max(cur_max,height[i]);
            result+=(cur_max-height[i]);
        }
        cur_max = height[height.size()-1];
        for(int i=height.size()-2;i>pos;i--)
        {
            cur_max = max(cur_max,height[i]);
            result+=(cur_max-height[i]);
        }
        return result;
    }
};

其实不需要找到最大值也可以。按照寻找最大值的思路。移动第二大的值。思路和上面一样。代码简洁些。

class Solution {
public:
    int trap(vector<int>& height) {
        if(height.size()<=2) return 0;
        int result=0;
        int left=0,right=height.size()-1,secHeight=0;
        while(left<right)
        {
            if(height[left]<height[right])
            {
                secHeight = max(height[left],secHeight);
                result+=(secHeight-height[left]);
                left++;
            }
            else
            {
                secHeight = max(height[right],secHeight);
                result+=(secHeight-height[right]);
                right--;
            }
        }
        return result;
    }
};