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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], 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!
思路：找到最大值及其位置pos。然后从左往右遍历到pos，再从右往左遍历到pos。
什么时候会加水呢？
遍历的时候记录最大值，也就是全局第二大的值，如果当前值比第二大的小，那么就会加水了，小多少就加多少。
加水是一列列加的。下图。
1 class Solution { 2 public: 3 int trap(vector<int>& height) { 4 if(height.size()<=2) return 0; 5 int result=0; 6 int pos=0,maxHeight=0; 7 for(int i=0;i<height.size();i++) 8 { 9 if(height[i]>maxHeight) 10 { 11 maxHeight = height[i]; 12 pos = i; 13 } 14 } 15 int cur_max = height[0]; 16 for(int i=1;i<pos;i++) 17 { 18 cur_max = max(cur_max,height[i]); 19 result+=(cur_max-height[i]); 20 } 21 cur_max = height[height.size()-1]; 22 for(int i=height.size()-2;i>pos;i--) 23 { 24 cur_max = max(cur_max,height[i]); 25 result+=(cur_max-height[i]); 26 } 27 return result; 28 } 29 };
其实不需要找到最大值也可以。按照寻找最大值的思路。移动第二大的值。思路和上面一样。代码简洁些。
1 class Solution { 2 public: 3 int trap(vector<int>& height) { 4 if(height.size()<=2) return 0; 5 int result=0; 6 int left=0,right=height.size()-1,secHeight=0; 7 while(left<right) 8 { 9 if(height[left]<height[right]) 10 { 11 secHeight = max(height[left],secHeight); 12 result+=(secHeight-height[left]); 13 left++; 14 } 15 else 16 { 17 secHeight = max(height[right],secHeight); 18 result+=(secHeight-height[right]); 19 right--; 20 } 21 } 22 return result; 23 } 24 };
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原文地址：http://www.cnblogs.com/Sean-le/p/4805186.html
