标签：style blog http io color os ar for sp
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!
找到最长的那块木板，假设其下标为maxIdx。
分别从左侧和右侧向其逼近。
在左侧逼近过程中：
如果一个木板的长度小于已经遍历的最大长度max，即max>该木板<maxIdx，所以在该木板位置能存max - 该木板长度的水量（左右两侧各有一个木板长于它）。
如果一个木板的长度大于已经遍历的最大长度max，即max<该木板<maxIdx，所以在该木板位置不能存水（因为左右两侧只有一个木板（maxIdx）长于它）。更新max值。
右侧逼近过程与左侧相似。
1 public class Solution { 2 public int trap(int[] A) { 3 if (A.length<3) { 4 return 0; 5 } 6 int len=A.length; 7 int maxIndex=0; 8 //找出最大值的下标 9 for (int i = 0; i < len; i++) { 10 if (A[i]>A[maxIndex]) { 11 maxIndex=i; 12 } 13 } 14 15 int leftMax=0; 16 int rightMax=0; 17 int water=0; 18 for (int i = 0; i < maxIndex; i++) { 19 if (leftMax>A[i]) { 20 water=water+leftMax-A[i]; 21 }else { 22 leftMax=A[i]; 23 } 24 } 25 26 for (int i = len-1; i > maxIndex; i--) { 27 if (rightMax>A[i]) { 28 water=water+rightMax-A[i]; 29 }else { 30 rightMax=A[i]; 31 } 32 } 33 return water; 34 } 35 }
标签：style blog http io color os ar for sp
原文地址：http://www.cnblogs.com/birdhack/p/4061002.html
