[LeetCode] Trapping Rain Water

最新推荐文章于 2021-11-04 09:56:06 发布

原创最新推荐文章于 2021-11-04 09:56:06 发布 · 274 阅读

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本文介绍了一种计算给定地形图中雨后积水体积的方法。通过分析地形高度数据，利用两次遍历策略，从左到右及从右到左确定每个位置上的积水高度，最终计算出总的积水体积。

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) {
        int max = A[0];
        int maxid = 0,sum = 0;
        for(int i = 1;i < n;i ++){
            if(A[maxid] < A[i])
                maxid = i;
        }
        for(int i = 1;i < maxid;i ++){
            if(A[i] >= max)
                max = A[i];
            else
                sum += max - A[i];
        }
        max = A[n - 1];
        for(int i = n - 2;i > maxid;i --){
            if(A[i] >= max)
                max = A[i];
            else
                sum += max - A[i];
        }
        return sum;
    }
};

参考