Trapping Rain Water (Java)

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!

從左到右遍歷一遍，再從右向左遍歷一遍，每次記錄當前i之前的最大值，作爲i的左邊最大值和右邊最大值。i的蓄水量只和Math.min(左邊最大值，右邊最大值) - A[i]有關

參考http://www.cnblogs.com/TenosDoIt/p/3812880.html

Source

    public int trap(int[] A) {
    	if(A.length == 0) return 0;
    	int len = A.length - 1;
    	int[] maxl = new int[len + 1];
    	int[] maxr = new int[len + 1];
    	int cap = 0, total = 0;
    	for(int i = len; i >= 0; i--){
    		maxr[i] = cap;
    		if(A[i] > cap) 
    			cap = A[i];
    	}
    	cap = 0;
    	for(int i = 0; i <= len; i++){
    		maxl[i] = cap;
    		if(A[i] > cap)
    			cap = A[i];
    	}
    	for(int i = 0; i <= len; i++){
    		int c = Math.min(maxl[i], maxr[i]) - A[i];
    		if(c > 0)   //*** -A[i]有可能是負數
    			total += c;
    	}
    	return total;
    }

Test

    public static void main(String[] args){
    	int[] A = {0, 1, 0, 2, 1, 0, 1, 3, 2, 1, 2, 1};
     	int b = new Solution().trap(A);
     	System.out.println(b);
     	
    }