LeetCode OJ-- Trapping Rain Water*

https://oj.leetcode.com/problems/trapping-rain-water/

模拟题,计算出在凹凸处存水量。

对于一个位置 i ,分别计算出它左边的最大值 left (从左扫描一遍), 右边的最大值 right(从右扫描一遍) 。找left right中的最小值,如果大于 A[i],则做 min - A[i] 的累加。

class Solution {
public:
    int trap(int A[], int n) {
        if(n<=1)
            return 0;

        vector<int> leftHigher;
        vector<int> rightHigher;
        int leftLarge = 0, rightLarge = 0;
        for(int i = 0;i<n;i++)
        {
            if(A[i]>leftLarge)
                leftLarge = A[i];
            leftHigher.push_back(leftLarge);
        }
        rightHigher.resize(n);
        for(int j = n-1;j>=0;j--)
        {
            if(A[j]>rightLarge)
                rightLarge = A[j];
            rightHigher[j] = rightLarge;
        }

        int sum = 0;
        for(int i = 0;i<n;i++)
        {
            int min = leftHigher[i]<rightHigher[i]?leftHigher[i]:rightHigher[i];
            if(min>A[i])
                sum += min - A[i];
        }
        
        return sum;
    }
};

 

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LeetCode OJ-- Trapping Rain Water*

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