Trapping Rain Water—LeetCode(2015/4/8腾讯基础研究线上笔试题)

原创于 2015-04-12 20:16:59 发布 · 471 阅读

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本文介绍了一种利用双指针算法解决计算下雨后能存留的雨水量问题的方法。通过一次遍历，找出每个位置上方的最高墙，从而计算出能存留的雨水量。

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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!

1、双指针，一次遍历

int trap(int A[], int n) {
        if(n<=2) return 0;
        int low=0;
        int high=n-1;
        
        int waterSum=0;
        int indexMin=0;
        
        while(low<high)
        {
            while(A[low]<=indexMin&&low<high)
            {
                waterSum+=(indexMin-A[low]);
                low++;
            }
            while(A[high]<=indexMin&&low<high)
            {
                waterSum+=(indexMin-A[high]);
                high--;
            }
            indexMin=A[low]>=A[high]?A[high]:A[low];
        }
        return waterSum;
    }

2、可以先找到序列中的最大值，再从两端向中间遍历，此时时间复杂度为O(2n);