Trapping Rain Water

描述

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.

Trapping Rain Water

Figure: Trapping Rain Water

分析

对于每个柱子，找到其左右两边最高的柱子，该柱子能容纳的面积就是min(max_left, max_right) - height。所以，

  • 从左往右扫描一遍，对于每个柱子，求取左边最大值；
  • 从右往左扫描一遍，对于每个柱子，求最大右值；

  • 再扫描一遍，把每个柱子的面积并累加。
    也可以，

  • 扫描一遍，找到最高的柱子，这个柱子将数组分为两半；

  • 处理左边一半；
  • 处理右边一半。

    代码1

  1. // Trapping Rain Water
  2. // 思路1，时间复杂度O(n)，空间复杂度O(n)
  3. class Solution {
  4. public:
  5.     int trap(const vector<int>& A) {
  6.         const int n = A.size();
  7.         int *left_peak = new int[n]();
  8.         int *right_peak = new int[n]();
  10.         for (int i = 1; i < n; i++) {
  11.             left_peak[i] = max(left_peak[i - 1], A[i - 1]);
  12.         }
  13.         for (int i = n - 2; i >=0; --i) {
  14.             right_peak[i] = max(right_peak[i+1], A[i+1]);
  15.         }
  17.         int sum = 0;
  18.         for (int i = 0; i < n; i++) {
  19.             int height = min(left_peak[i], right_peak[i]);
  20.             if (height > A[i]) {
  21.                 sum += height - A[i];
  22.             }
  23.         }
  25.         delete[] left_peak;
  26.         delete[] right_peak;
  27.         return sum;
  28.     }
  29. };

代码2

  1. // Trapping Rain Water
  2. // 思路2，时间复杂度O(n)，空间复杂度O(1)
  3. class Solution {
  4. public:
  5.     int trap(const vector<int>& A) {
  6.         const int n = A.size();
  7.         int peak_index = 0; // 最高的柱子，将数组分为两半
  8.         for (int i = 0; i < n; i++)
  9.             if (A[i] > A[peak_index]) peak_index = i;
  11.         int water = 0;
  12.         for (int i = 0, left_peak = 0; i < peak_index; i++) {
  13.             if (A[i] > left_peak) left_peak = A[i];
  14.             else water += left_peak - A[i];
  15.         }
  16.         for (int i = n - 1, right_peak = 0; i > peak_index; i--) {
  17.             if (A[i] > right_peak) right_peak = A[i];
  18.             else water += right_peak - A[i];
  19.         }
  20.         return water;
  21.     }
  22. };

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原文: https://soulmachine.gitbooks.io/algorithm-essentials/content/cpp/linear-list/array/trapping-rain-water.html