[LeetCode] 接雨水,题 Trapping Rain Water

这题放上来是因为自己第一回见到这种题,觉得它好玩儿 =)

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!

class Solution {

public:

    int trap(int A[], int n) {

    }

};

 

题目不难,观察下就可以发现被水填满后的形状是先升后降的塔形,因此,先遍历一遍找到塔顶,然后分别从两边开始,往塔顶所在位置遍历,水位只会增高不会减小,且一直和最近遇到的最大高度持平,这样知道了实时水位,就可以边遍历边计算面积。

从看题目开始一共十分钟,算本菜鸟在AC率二字打头的题目中至今最快的一次。。

class Solution {

public:

    int trap(int A[], int n) {

        if(n <= 2) return 0;

        int max = -1, maxInd = 0;

        int i = 0;

        for(; i < n; ++i){

            if(A[i] > max){

                max = A[i];

                maxInd = i;

            }

        }

        int area = 0, root = A[0];

        for(i = 0; i < maxInd; ++i){

            if(root < A[i]) root = A[i];

            else area += (root - A[i]);

        }

        for(i = n-1, root = A[n-1]; i > maxInd; --i){

            if(root < A[i]) root = A[i];

            else area += (root - A[i]);

        }

        return area;

    }

};

 

 

总结:

这道题和LeetCode上 "Candy" 一题都采用了定义两个指针向中部某一点靠拢的做法(当然Candy还有更快的解,见以前的这篇),这也算是小技巧之一吧,在需要时要能第一时间想到。

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