LeetCode 111 Trapping Rain 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.

For example,
Given [0,1,0,2,1,0,1,3,2,1,2,1], return 6.



分析：

对于每一格A[i]，计算左边的最高沿left[i]，再计算右边的最高沿right[i]，根据木桶原理，这一格能存的水就是：

vol = min{left[i], right[i]} - A[i] > 0 ? min{left[i], right[i]} - A[i] : 0;

public class Solution {
public int trap(int[] A) {
if(A==null || A.length==0)
return 0;
int[] left = new int[A.length];
int[] right = new int[A.length];
//计算左边最大
int max = A[0];
for(int i=0; i<A.length; i++){
if(A[i] <= max){
left[i] = max;
}else{
left[i] = A[i];
max = A[i];
}
}
//计算右边最大
max=A[A.length-1];
for(int i=A.length-1; i>=0; i--){
if(A[i] <= max){
right[i] = max;
}else{
right[i] = A[i];
max = A[i];
}
}
//计算储水，0和A.length-1位置不算
int vol = 0;
for(int i=1; i<A.length-1; i++){
//左右沿取最低
int height = Math.min(left[i], right[i]);
if(height > A[i]){
vol += height-A[i];
}
}
return vol;
}
}