Trapping Rain Water

Problem:

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.

    Solution1是自己首先想到的一个版本，遇到下降的块压栈，遇到上升的块出栈，计算矩形面积。时间复杂度为O(n)，空间复杂度为O(n)。

    Solution2参考了别人的解决方法。想法就是先计算总面积，再减去实块所占的面积。时间复杂度O(n)，空间复杂度O(1)。

Solution1:

public class Solution {

    public int trap(int[] A) {
        if(A==null||A.length==0)
             return 0;
        Stack<Integer> st = new Stack<>();
        int water = 0;
        int pre = 0;
        for(int i=0;i<A.length;i++)
        {
             if(!st.empty()&&(A[i]>A[st.peek()]||i>st.peek()+1))
             {
                 int j=st.peek();
                 while(!st.empty()&&A[i]>=A[st.peek()])
                 {
                     j = st.pop();
                     water += (i-j-1)*(A[j]-pre);
                     pre = A[j];
                 }
                 if(!st.empty())
                 {
                     water += (i-st.peek()-1)*(A[i]-pre);
                     pre = A[i];
                 }
             }
             else
                 pre = 0;
             st.push(i);
        }
        return water;
    }

}

Solution2:

public class Solution {
     public int trap(int[] A){
    if(A==null||A.length<=1)
        return 0;
    int blocks = 0,all = 0,b = 0,e = A.length-1,height = 0;
    while(b<e)
    {
         while(b<e&&A[b]<=height)blocks += A[b++];
        while(b<e&&A[e]<=height)blocks += A[e--];
        all += (Math.min(A[b], A[e])-height)*(e-b+1);
         height = Math.min(A[b], A[e]);
    }
     return all - blocks - A[b];
 }
}