[LeetCode]Trapping Rain Water,解题报告

前言

在LeetCode上又碰到了一道笔试原题，充分说明提前的准备对应届生找工作有多大的作用，年终总结我会具体写一下自己找工作的准备历程，耗时将近半年时间

题目

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.

解题思路

首先，碰到这样的题目不要慌张，挨个分析每个A[i]能trapped water的容量，然后将所有的A[i]的trapped water容量相加即可

其次，对于每个A[i]能trapped water的容量，取决于A[i]左右两边的高度（可延展）较小值与A[i]的差值，即volume[i] = [min(left[i], right[i]) - A[i]] * 1，这里的1是宽度，如果the width of each bar is 2,那就要乘以2了

AC代码

public class Solution {
public int trap(int[] A) {
// special case
if (A == null || A.length == 0) {
return 0;
}

int i, max, volume, left[] = new int[A.length], right[] = new int[A.length];

// from left to right
for (left[0] = A[0], i = 1, max = A[0]; i < A.length; i++) {
if (A[i] < max) {
left[i] = max;
} else {
left[i] = A[i];
max = A[i];
}
}

// from right to left
for (right[A.length - 1] = A[A.length - 1], i = A.length - 2, max = A[A.length - 1]; i >= 0; i--) {
if (A[i] < max) {
right[i] = max;
} else {
right[i] = A[i];
max = A[i];
}
}

// trapped water
for (volume = 0, i = 1; i <= A.length - 2; i++) {
int tmp = Math.min(left[i], right[i]) - A[i];
if (tmp > 0) {
volume += tmp;
}
}

return volume;
}
}