605. Can Place Flowers*

​​https://leetcode.com/problems/can-place-flowers/​​

题目描述

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:

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

C++ 实现 1

贪婪算法. 能用上最少空间的思路是, 遇到为 ​​0​​​ 的位置, 就尝试放置一盆花, 但是要判断左边 ​​prev​​ 以及右边是否已经有花了.

思路参考: ​​Java - Greedy solution - O(flowerbed) - beats 100%​​

其中 ​​count < n​​ 这个条件是一个优化, 可以提前终止循环.

class Solution {
public:
    bool canPlaceFlowers(vector<int>& flowerbed, int n) {
        int count = 0;
        // 利用 count < n 提前终止循环
        for (int i = 0; i < flowerbed.size() && count < n; ++ i) {
            if (flowerbed[i] == 0) {
                int prev = i == 0 ? 0 : flowerbed[i - 1];
                int next = i == flowerbed.size() - 1 ? 0 : flowerbed[i + 1];
                if (prev == 0 && next == 0) {
                    flowerbed[i] = 1;
                    count ++;
                }
            }
        }
        return count == n;
    }
};