leetcode 605. Can Place Flowers

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.

我的方法是枚举

class Solution {
public:
bool canPlaceFlowers(vector<int>& flowerbed, int n) {
int len = flowerbed.size();
int num = 0;
for (int i = 0; i < len; i +=2) {
if (flowerbed[i] == 0) {
if ( (i+1 >=len ||flowerbed[i + 1] == 0) && (i-1 < 0 ||flowerbed[i - 1] == 0) ) num++;
}
}
if (num >= n) return true;
num = 0;
for (int i = 1; i < len; i += 2) {
if (flowerbed[i] == 0) {
if ((i+1 >= len || flowerbed[i + 1] ==0) && (i-1 < 0 || flowerbed[i - 1] == 0)) num++;
}
}
if (num >= n) return true;
return false;
}
};

网上看了个不错的思路如下：

class Solution {
public:
bool canPlaceFlowers(vector<int>& flowerbed, int n) {
int result = 0;
int count = 1;
// edge case: left and right
for (int i = 0; i < flowerbed.size(); i++) {
if (flowerbed[i] == 0) {
count++;
} else {
result += (count - 1) / 2;
count = 0;
}
}
if (count != 0) {
result += count / 2;
}
return result >= n;
}
};