LeetCode——Trapping Rain Water

标签：leetcode

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!

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

分析：类似于木桶原理，它能存储的水取决于一个区间内最左侧的最高的一格和右侧最高的一格。所以先找到最高的点、小于最高点的左右次高点，保存左右到最高点间的水量。

	public static int trap(int[] A){
		int len = A.length;
		if(len <= 2)
			return 0;
		int higest = A[0],high_index = 0;
		for(int i=0;i<len;i++){
			int tmp = A[i];
			if(higest < tmp){
				higest = A[i];
				high_index = i;
			}
		}
		int sum = 0;
		int maxl = A[0],maxr = A[len-1];
		for(int i=0;i<high_index;i++){
			if(A[i] > maxl)
				maxl = A[i];
			else
				sum += maxl - A[i];
		}
		for(int i=len-1;i>high_index;i--){
			if(A[i] >maxr)
				maxr = A[i];
			else
				sum += maxr-A[i];
		}
		return sum;
	}

LeetCode——Trapping Rain Water

标签：leetcode

原文地址：http://blog.csdn.net/laozhaokun/article/details/39254709