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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.
Given [0,1,0,2,1,0,1,3,2,1,2,1]
, return 6
.
O(n) time and O(1) memory
O(n) time and O(n) memory is also acceptable.
public class Solution { /** * @param heights: an array of integers * @return: a integer */ public int trapRainWater(int[] heights) { // write your code here if(heights == null || heights.length == 0) return 0; int[] left = new int[heights.length]; int[] right = new int[heights.length]; int res = 0; left[0] = 0; right[heights.length - 1] = 0; for(int i = 1; i < heights.length; i++) left[i] = Math.max(heights[i - 1], left[i - 1]); for(int i = heights.length - 2; i >= 0; i--) right[i] = Math.max(heights[i + 1], right[i + 1]); for(int i = 0; i < heights.length; i++){ int height = Math.min(left[i], right[i]); if(height > heights[i]) res += height - heights[i]; } return res; } }
lintcode-medium-Trapping Rain Water
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原文地址:http://www.cnblogs.com/goblinengineer/p/5362953.html