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!

也是一个波的问题。

public class Solution {
    public int trap(int[] A) {
        // Note: The Solution object is instantiated only once and is reused by each test case.
        int left = -1;
        int right = -1;
        int sum = 0;
        int i = 0;
        for(; i < A.length; i ++){
            if(left == -1 && A[i] != 0){
                left = i;
                break;
            }
        }
        for(i = A.length - 1; i >= 0; i --){
            if(right == -1 && A[i] != 0){
                right = i;
                break;
            }
        }
        
        if(left != -1){
            i = left + 1;
            while(i < A.length){
                if(A[left] <= A[i]){
                    for(int j = left; j < i; j ++){
                        sum += A[left] - A[j];
                    }
                    left = i;
                }
                i ++;
            }
        }
        
        
        if(right != -1){
            i = right - 1;
            while(i > -1){
                if(A[right] < A[i]){
                    for(int j = right; j > i; j --){
                        sum += A[right] - A[j];
                    }
                    right = i;
                }
                i --;
            }
        }
        return sum;
    }
}

 第二遍：

public class Solution {
    public int trap(int[] A) {
        // Note: The Solution object is instantiated only once and is reused by each test case.
        int n = A.length;
        int[] b = new int[n];
        int M=-10000;
        for (int i=0; i<n; i++){
            b[i] = M;
            M = Math.max( M, A[i] );
        }
        M=-10000;
        int ret=0;
        for (int i=n-1; i>=0; i--){
            if ( i!=n-1 && i!=0 ){
                if ( A[i] < Math.min( M, b[i] ) )
                    ret += Math.min( M, b[i] )-A[i];
            }
            M = Math.max( M, A[i] );
        }
        return ret;
    }
}