[leetcode] Trapping Rain Water

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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], return6.

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!

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

思路1：對某個值A[i]來說，能裝的最多的水取決於在i之前最高的值leftMaxt[i]和在i右邊的最高的rightMax[i]，容水量即min(left,right) – A[i]。:為了計算高度，第一遍從左至右計算數組leftMaxt，第二遍從右至左計算rightMax。時空複雜度都是O(n)。

思路2：改進思路1，可以用常量空間搞定。具體思路，先找到最高的bar，然後總兩邊向中間計算，邊計算邊更新curHeight。

具體可參考 這裡

思路1：

public class Solution {public int trap(int[] A) {if (A == null || A.length < 3)return 0;int n = A.length;int[] leftMax = new int[n];int[] rightMax = new int[n];int i;int max = A[0];for (i = 1; i < n - 1; i++) {leftMax[i] = max;if (A[i] > max)max = A[i];}max = A[n - 1];for (i = n - 2; i > 0; i--) {rightMax[i] = max;if (A[i] > max)max = A[i];}int count = 0;for (i = 1; i < n - 1; i++) {int min = leftMax[i] < rightMax[i] ? leftMax[i] : rightMax[i];if (min > A[i])count += min - A[i];}return count;}public static void main(String[] args) {System.out.println(new Solution().trap(new int[] { 0, 1, 0, 2, 1, 0, 1,3, 2, 1, 2, 1 }));}}

 

 

 

參考：

http://fisherlei.blogspot.com/2013/01/leetcode-trapping-rain-water.html

http://www.cnblogs.com/lichen782/p/Leetcode_Trapping_Rain_Water.html