LeetCode: Trapping Rain Water 解题报告

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

Trapping Rain Water
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!

SOLUTION 1:

从左到右扫描，计算到从左边到curr的最高的bar,从右到左扫描，计算到从右边到curr的最高的bar。

再扫描一次，把这两者的低者作为{桶}的高度，如果这个桶高于A[i]的bar,那么A[i]这个bar上头可以存储height - A[i]这么多水。把这所有的水加起来即可。

 public class Solution {
     public int trap(int[] A) {
         if (A == null) {
             return ;
         }

         int max = ;

         int len = A.length;
         int[] left = new int[len];
         int[] right = new int[len];

         // count the highest bar from the left to the current.
         for (int i = ; i < len; i++) {
             left[i] = i ==  ? A[i]: Math.max(left[i - ], A[i]);
         }

         // count the highest bar from right to current.
         for (int i = len - ; i >= ; i--) {
             right[i] = i == len -  ? A[i]: Math.max(right[i + ], A[i]);
         }

         // count the largest water which can contain.
         for (int i = ; i < len; i++) {
             int height = Math.min(right[i], left[i]);
             if (height > A[i]) {
                 max += height - A[i];
             }
         }

         return max;
     }
 }

2015.1.14 redo:

合并2个for 循环，简化计算。

 public class Solution {
     public int trap(int[] A) {
         // 2:37
         if (A == null) {
             return ;
         }

         int len = A.length;
         int[] l = new int[len];
         int[] r = new int[len];

         for (int i = ; i < len; i++) {
             if (i == ) {
                 l[i] = A[i];
             } else {
                 l[i] = Math.max(l[i - ], A[i]);
             }
         }

         int water = ;
         for (int i = len - ; i >= ; i--) {
             if (i == len - ) {
                 r[i] = A[i];
             } else {
                 // but: use Math, not max
                 r[i] = Math.max(r[i + ], A[i]);
             }

             water += Math.min(l[i], r[i]) - A[i];
         }

         return water;
     }
 }

CODE:

https://github.com/yuzhangcmu/LeetCode_algorithm/blob/master/array/Trap.java