凹槽最大水容量

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

思路：

对某个值A[i]来说，能trapped的最多的water取决于在i之前最高的值leftMostHeight[i]和在i右边的最高的值rightMostHeight[i]（均不包含自身）。

如果min(left,right) > A[i]，那么在i这个位置上能trapped的water就是min(left,right) – A[i]。

有了这个想法就好办了，第一遍从左到右计算数组leftMostHeight，第二遍从右到左计算rightMostHeight。
时间复杂度是O(n)。

public class Solution {
     public int trap(int[] A) {
if(A == null || A.length < 1)
            return 0;  
          
        int maxheight = 0; 
        int[] leftMostHeight = new int[A.length];  
        for(int i =0; i < A.length; i++)  {  
            leftMostHeight[i] = maxheight;  
            maxheight = maxheight > A[i] ? maxheight : A[i];  
        }  
  
        maxheight = 0;  
       int[] rightMostHeight = new int[A.length];  
        for(int i =A.length - 1; i>=0; i--) {  
            rightMostHeight[i] = maxheight;  
            maxheight = maxheight > A[i] ? maxheight : A[i];  
        }  
  
        int water = 0;  
        for(int i =0; i < A.length; i++)  
        {  
            int high = Math.min(leftMostHeight[i], rightMostHeight[i]) - A[i];  
            if(high>0)  
                water += high;  
        }  
        return water;

}

}

解法二：

是两边往中间遍历，
记录当前第二高点secHight，
然后利用这个第二高点减去当前历经的柱子，
剩下就装水容量了；
两边比较时，最高的点不用动，只移动第二高点  
    public class Solution {
     public int trap(int[] A) {    

 int secHight = 0;
int left = 0;
int right =A.length;
int area = 0;
while (left < right){
if (A[left] < A[right]){
secHight = Math.max(A[left], secHight);
area += secHight-A[left];//计算当前格的能装雨水的容量
left++;
} else {
secHight = Math.max(A[right], secHight);
area += secHight-A[right];
right--;
}
}
return area;

}

}