LeetCode: Trapping Rain Water 解题报告

https://oj.leetcode.com/problems/trapping-rain-water/

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!

SOLUTION 1:

从左到右扫描，计算到从左边到curr的最高的bar,从右到左扫描，计算到从右边到curr的最高的bar。

再扫描一次，把这两者的低者作为{桶}的高度，如果这个桶高于A[i]的bar,那么A[i]这个bar上头可以存储height - A[i]这么多水。把这所有的水加起来即可。

public class Solution {
    public int trap(int[] A) {
        if (A == null) {
            return 0;
        }
        
        int max = 0;
        
        int len = A.length;
        int[] left = new int[len];
        int[] right = new int[len];
        
        // count the highest bar from the left to the current.
        for (int i = 0; i < len; i++) {
            left[i] = i == 0 ? A[i]: Math.max(left[i - 1], A[i]);
        }
        
        // count the highest bar from right to current.
        for (int i = len - 1; i >= 0; i--) {
            right[i] = i == len - 1 ? A[i]: Math.max(right[i + 1], A[i]);
        }
        
        // count the largest water which can contain.
        for (int i = 0; i < len; i++) {
            int height = Math.min(right[i], left[i]);
            if (height > A[i]) {
                max += height - A[i];
            }
        }
        
        return max;
    }
}

2015.1.14 redo:

合并2个for 循环，简化计算。

public class Solution {
    public int trap(int[] A) {
        // 2:37
        if (A == null) {
            return 0;
        }
        
        int len = A.length;
        int[] l = new int[len];
        int[] r = new int[len];
        
        for (int i = 0; i < len; i++) {
            if (i == 0) {
                l[i] = A[i];
            } else {
                l[i] = Math.max(l[i - 1], A[i]);
            }
        }
        
        int water = 0;
        for (int i = len - 1; i >= 0; i--) {
            if (i == len - 1) {
                r[i] = A[i];
            } else {
                // but: use Math, not max
                r[i] = Math.max(r[i + 1], A[i]);
            }
            
            water += Math.min(l[i], r[i]) - A[i];
        }
        
        return water;        
    }
}

CODE:

https://github.com/yuzhangcmu/LeetCode_algorithm/blob/master/array/Trap.java