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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:
- 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.
题意:给定一个只包含0和1的一位数组,0表示这个位置是空的,1表示这个位置上已经种植了花;现在给定还需要种植的花数n,要求两朵花之间的最小距离为2,即种植花的位置的左右相邻位置均不能够种植花,判断是否还能种植给定花数;
解法:我们只需要在遍历数组的时候,每当遇到一个0,就判断其是否可以种植,满足左右相邻位置均为空即可;若能够种植,将此位置置为1,并且将n值减1,否则继续寻找下一个0,在遍历的过程中,当n值减为1时则返回true;
class Solution {
public boolean canPlaceFlowers(int[] flowerbed, int n) {
if (n == 0) {
return true;
}
for (int i = 0; i < flowerbed.length; i++) {
if (flowerbed[i] == 0 && (i == 0 || flowerbed[i - 1] != 1) &&
(i == flowerbed.length - 1 || flowerbed[i + 1] != 1)) {
n--;
flowerbed[i] = 1;
}
if (n == 0) {
return true;
}
}
return false;
}
}