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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。
1 public class Solution { 2 public int trap(int[] A) { 3 if (A == null) { 4 return 0; 5 } 6 7 int max = 0; 8 9 int len = A.length;10 int[] left = new int[len];11 int[] right = new int[len];12 13 // count the highest bar from the left to the current.14 for (int i = 0; i < len; i++) {15 left[i] = i == 0 ? A[i]: Math.max(left[i - 1], A[i]);16 }17 18 // count the highest bar from right to current.19 for (int i = len - 1; i >= 0; i--) {20 right[i] = i == len - 1 ? A[i]: Math.max(right[i + 1], A[i]);21 }22 23 // count the largest water which can contain.24 for (int i = 0; i < len; i++) {25 int height = Math.min(right[i], left[i]);26 if (height > A[i]) {27 max += height - A[i];28 }29 }30 31 return max;32 }33 }
CODE:
https://github.com/yuzhangcmu/LeetCode_algorithm/blob/master/array/Trap.java
LeetCode: Trapping Rain Water 解题报告
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