LEETCODE: Trapping Rain Water

最新推荐文章于 2019-07-04 18:17:10 发布

原创最新推荐文章于 2019-07-04 18:17:10 发布 · 342 阅读

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#LEETCODE #算法 #数组

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本文介绍了一个算法，用于计算给定高度数组表示的地形中，在雨后能够储存的雨水量。通过遍历数组并确定最高点，算法能够有效计算雨水储存量。

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 maxi = 0;
        for(int ii = 1; ii < n; ii ++) {
            if(A[maxi] < A[ii]) {
                maxi = ii;
            }
        }
        
        int result = 0;
        // Left to max
        int maxval = A[0];
        for(int ii = 1; ii < maxi; ii ++) {
            if(A[ii] > maxval) {
                maxval = A[ii];
            }
            else {
                result += maxval - A[ii];
            }
        }
        
        // Right to max
        maxval = A[n - 1];
        for(int ii = n -2; ii > maxi; ii --) {
            if(A[ii] > maxval) {
                maxval = A[ii];
            }
            else {
                result += maxval - A[ii];
            }
        }
        
        return result;
    }
};