标签：set += array load ++ cas trap app water
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.
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!
Example:
Input: [0,1,0,2,1,0,1,3,2,1,2,1] Output: 6
class Solution { public int trap(int[] A){ int a=0; int b=A.length-1; int max=0; int leftmax=0; int rightmax=0; while(a<=b){ leftmax=Math.max(leftmax,A[a]); rightmax=Math.max(rightmax,A[b]); if(leftmax<rightmax){ max+=(leftmax-A[a]); // leftmax is smaller than rightmax, so the (leftmax-A[a]) water can be stored a++; } else{ max+=(rightmax-A[b]); b--; } } return max; } }
leetcode 42. Trapping Rain Water
标签：set += array load ++ cas trap app water
原文地址：https://www.cnblogs.com/jamieliu/p/10347244.html
