leetcode Trapping Rain Water

最新推荐文章于 2020-08-24 17:46:40 发布

转载最新推荐文章于 2020-08-24 17:46:40 发布 · 152 阅读

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原文链接：http://www.cnblogs.com/riasky/p/3455550.html

本文介绍了一种计算给定高度地图上积水体积的方法。通过寻找最高点，然后分别从两端向最高点遍历，计算每个位置能容纳的雨水量。此算法巧妙地解决了经典的“接雨水”问题。

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Trapping Rain Water

Total Accepted: 2335 Total Submissions: 8464 My Submissions

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!

Discuss

Find the highest, then from the start to the highest, then the last to the highest..

class Solution {
 public:
  int trap(int A[], int n) {
    if (n <= 2)
      return 0;
    int i, maxElevation = A[0], maxIndex = 0, h = 0, res = 0;
    for (i = 1; i < n; ++i)
      if (A[i] > maxElevation) {
        maxElevation = A[i];
        maxIndex = i;
      }
    for (i = 0; i <= maxIndex - 1; ++i) 
      if (A[i] >= h)
        h = A[i];
      else
        res += (h - A[i]);
    
    h = 0;
    for (i = n - 1; i >= maxIndex + 1; --i) 
      if (A[i] >= h)
        h = A[i];
      else
        res += (h - A[i]);
    return res;
  }
};

转载于:https://www.cnblogs.com/riasky/p/3455550.html