[LeetCode] 605. Can Place Flowers

轉自LeetCode

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.

<Solution>

想法如下

• 歷遍 array，並根據條件檢查可種花的位置有幾個

• 如果已經滿足條件了，就可以直接回傳答案，不用再做檢查

code 如下

Java

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

view raw canPlaceFlowers.cpp hosted with ❤ by GitHub

kotlin

	class Solution {
	fun canPlaceFlowers(flowerbed: IntArray, n: Int): Boolean {
	return if(n == 0) {
	true
	} else {
	var next = 0
	var prv = 0
	var count = 0
	for(i in flowerbed.indices) {
	if(flowerbed[i] == 0) {
	next = if(i == flowerbed.lastIndex) {
	0
	} else {
	flowerbed[i+1]
	}
	prv = if(i == 0) {
	0
	} else {
	flowerbed[i-1]
	}
	if(next + prv == 0) {
	flowerbed[i] = 1
	++count
	}
	}
	}
	count >= n
	}
	}
	}

view raw can_place_flowers.kt hosted with ❤ by GitHub