Leetcode: 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!

这道题做的比较艰辛，一开始自己想的是一个用stack的解法，感觉过于繁琐（出栈，入栈，计算容积），但未尝不是一个好的尝试，这个方法还是有点小问题，过后会好好想清楚。看了网上提示完成了最终的方法，这个方法两次遍历数组，第一次遍历找每个元素右边最大的元素，第二次遍历寻找每个元素左边最大的元素，同时计算该index可以存放的水容积： min{lefthighest, righthighest}-A[i]

用两个array

public class Solution {
    public int trap(int[] A) {
        int[] left = new int[A.length];
        int[] right = new int[A.length];
        int sum = 0;
        for (int i=0; i<A.length; i++) {
            if (i == 0) left[i] = 0;
            else {
                left[i] = Math.max(left[i-1], A[i-1]);
            }
        }
        for (int i=A.length-1; i>=0; i--) {
            if (i == A.length - 1) right[i] = 0;
            else {
                right[i] = Math.max(right[i+1], A[i+1]);
            }
            if (Math.min(left[i], right[i]) > A[i]) {
                sum += Math.min(left[i], right[i]) - A[i];
            }
        }
        return sum;
    }
}

用一个array和一个变量

public class Solution {
    public int trap(int[] A) {
        int[] left = new int[A.length];
        int right = 0;
        int sum = 0;
        for (int i=0; i<A.length; i++) {
            if (i == 0) left[i] = 0;
            else {
                left[i] = Math.max(left[i-1], A[i-1]);
            }
        }
        for (int i=A.length-1; i>=0; i--) {
            if (i == A.length - 1) right = 0;
            else {
                right = Math.max(right, A[i+1]);
            }
            if (Math.min(left[i], right) > A[i]) {
                sum += Math.min(left[i], right) - A[i];
            }
        }
        return sum;
    }
}