LeetCode 605. Can Place Flowers

问题描述

Suppose you have a long flowerbed in which some of the plots are planted and some are not. However, flowers cannot be planted in adjacent plots - they would compete for water and both would die.

Given a flowerbed (represented as an array containing 0 and 1, where 0 means empty and 1 means not empty), and a number n, return if n new flowers can be planted in it without violating the no-adjacent-flowers rule.
Example 1:

Input: flowerbed = [1,0,0,0,1], n = 1
Output: True

Example 2:

Input: flowerbed = [1,0,0,0,1], n = 2
Output: False

Note:

1. The input array won’t violate no-adjacent-flowers rule.
2. The input array size is in the range of [1, 20000].
3. n is a non-negative integer which won’t exceed the input array size.

问题分析

给定一个数组nums[],数组中有一些0，1.其中0，表示空，1表示非空。给定一个数n，表示n给1，将这n个1插入到nums中，要求相邻两个不能是1.返回结果表示成功与否。
遍历这个数组，满足条件，计数+1.同时跳两位，如果不满足，就跳一位。时间复杂度 O(n).

代码实现

public boolean canPlaceFlowers(int[] flowerbed, int n) {     if (n == 0) {            return true;        }        if (flowerbed.length < (2 * n - 1)) {            return false;        }        int i = 0;        int count = 0;        while (i < flowerbed.length) {            if (flowerbed[i] == 0 && (i == 0 || flowerbed[i - 1] == 0) && (i == flowerbed.length - 1 || flowerbed[i + 1] == 0)) {                flowerbed[i] = 0;                count++;                i++;            }            if (count >= n) {                return true;            }            i++;        }        return false;    }

总结

在思考的时候，尽可能使用统一 的循环，而不用对其中不同的情况进行列举，那样容易出错。