【Leetcode】605. Can Place Flowers

标签：a long   continue   cannot   nta   ber   with   pos   des   ever   

Description

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.

Discuss

只需要计算出一共可以放多少个就可以了。

Code

class Solution {
    public boolean canPlaceFlowers(int[] flowerbed, int n) {
        int len = flowerbed.length;
        if (len < n) { return false; }
        int count = 0;
        if (len == 1 && flowerbed[0] == 0) { 
            count++; 
            return true;
        }
        for (int i = 0; i < len; i++) {
            if (i == 0) {
                if (flowerbed[0] == 0 && flowerbed[1] == 0) {
                    count++;
                    flowerbed[0] = 1;
                }
                continue;
            }
            if (i == len-1) {
                if (flowerbed[i - 1] == 0 && flowerbed[i] == 0) {
                    count++;
                }
                break;
            }
            if (flowerbed[i - 1] == 0 && flowerbed[i] == 0 && flowerbed[i + 1] == 0) {
                count++;
                flowerbed[i] = 1;
            }
        }
        return count >= n;
    }
}

【Leetcode】605. Can Place Flowers

标签：a long   continue   cannot   nta   ber   with   pos   des   ever   

原文地址：https://www.cnblogs.com/xiagnming/p/9389911.html