LeetCode-042：Trapping Rain Water

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题目：

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

题意：

求雨水体积，看图就明白了

思路：

一开始暴力，380ms，妈耶，太久了，不好~然后设置l、r两个指针从首尾两边遍历，这样能节省时间，在当前两指针确定的范围内，先比较两头找出较小值，如果较小值是l指向的值，则从左向右扫描，如果较小值是r指向的值，则从右向左扫描，若遇到的值比当较小值小，则将差值存入结果，如遇到的值大，则重新确定新的窗口范围，以此类推直至l==r

Code：

解法一：

class Solution {
public:
    int trap(vector<int>& height) {
        int i,j,k,len=height.size(),num=0;
        for(i=1;i<len;i++){
            int maxl=0,maxr=0;
            j=k=i;
            while(j>=0) maxl=max(maxl,height[j]),j--;
            while(k<len) maxr=max(maxr,height[k]),k++;
            num+=min(maxl,maxr)-height[i];
        }
        return num;
    }
};

解法二：

class Solution {
public:
    int trap(vector<int>& height) {
        int num=0,l=0,r=height.size()-1;
        while(l<r){
            int minr=min(height[l],height[r]);
            if(minr==height[l]){
                l++;
                while(l<r&&height[l]<minr){
                    num+=minr-height[l++];
                }
            }
            else{
                r--;
                while(l<r&&height[r]<minr){
                    num+=minr-height[r--];
                }
            }
        }
        return num;
    }
};