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.

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!

 第一遍：

public class Solution {
    public int trap(int[] A) {
        if(A == null || A.length < 3) return 0;
        int start = 0, end = A.length - 1;
        while(A[start] == 0){
            start ++;
        }
        while(A[end] == 0){
            end --;
        }
        int left = start;
        int sum = 0;
        for(int i = start + 1; i <= end; i ++){
            if(A[i] >= A[left]){
                for(int j = left + 1; j < i; j ++){
                    sum += A[left] - A[j];
                }
                left = i;
            }
        }
        int right = end;
        for(int i = end - 1; i >= start; i --){
            if(A[i] > A[right]){
                for(int j = i + 1; j < right; j ++){
                    sum += A[right] - A[j];
                }
                right = i;
            }
        }
        return sum;
    }
}

第三遍：

public class Solution {
    public int trap(int[] A) {
        if(A == null || A.length < 3) return 0;
        int sum = 0;
        int start = 0, end = A.length - 1;
        while(start < A.length && A[start] == 0) start ++;
        while(end > -1 && A[end] == 0) end --;
        int per = start, cur = start + 1;
        for(; cur <= end; cur ++){
            if(A[cur] >= A[per]){
                for(int i = per + 1; i < cur; i ++)
                    sum += A[per] - A[i];
                per = cur;
            }
        }
        per = end; cur = end - 1;
        for(; cur >= start; cur --){
            if(A[cur] > A[per]){
                for(int i = cur + 1; i < per; i ++)
                    sum += A[per] - A[i];
                per = cur;
            }
        }
        return sum;
    }
}