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 nnew 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.
枚举法写出来 答案,计算可以种花总数是否比n大即可,有投机的成分
//如果这个东西是奇偶数,如果里面的客房偶数比b大即可 /* 左为0 当前为0 且右边为0或边界 这时改为1 count++ 左为0 当前为1 继续 左为1 当前为0 继续 左为1 当前为1 继续 */ public class Solution { public boolean canPlaceFlowers(int[] flowerbed, int n) { int count =0; int len=flowerbed.length; if(len==1&&flowerbed[0]==0){ count++; } if(len==2&&(flowerbed[0]==0&&flowerbed[1]==0)){ count++; } for(int i=1;len>1&&i<len-1;i++){ if(i==1&&flowerbed[i]==0&&flowerbed[i-1]==0){ flowerbed[i-1] =1; count++; } if((flowerbed[i]==0)&&(flowerbed[i]==flowerbed[i-1])&&(flowerbed[i]==flowerbed[i+1])){ flowerbed[i]=1; count++; } if(i==len-2&&flowerbed[i]==0&&flowerbed[i+1]==0){ flowerbed[i+1] =1; count++; } } if(count>=n) {return true;} return false; } }
第二种办法,根据一次写的边界值进行归类
因为只需要判断当前元素如果是0的时候与左、右边界值(如果存在着)进行对比
//如果这个东西是奇偶数,如果里面的客房偶数比b大即可 /* 左为0 当前为0 且右边为0或边界 这时改为1 count++ 左为0 当前为1 继续 左为1 当前为0 继续 左为1 当前为1 继续 */ public class Solution { public boolean canPlaceFlowers(int[] flowerbed, int n) { int count =0; int len=flowerbed.length; //boolean flag =flase; int right=-1; int left =-1; for(int i=0;i<len;i++){ //判断当前是否为0,如果为0,且右边有值则比较 否则直接改 int mid=flowerbed[i]; if(mid==0){ //右边界 if(i==0){ if(i+1<len){ right =flowerbed[i+1]; if(mid==right){ flowerbed[i]=1; count++; } } else { flowerbed[i]=1; count++; } } //右边界 if(i==len-1){ if(i-1>=0){ left =flowerbed[i-1]; if(mid==left){ flowerbed[i]=1; count++; } } else { flowerbed[i]=1; count++; } } //中间值 if(i-1>=0&&i+1<len){ left =flowerbed[i-1]; right =flowerbed[i+1]; if(left==mid&&mid==right){ flowerbed[i]=1; count++; } } } } if(count>=n) {return true;} return false; } }
官方答案1
public class Solution { public boolean canPlaceFlowers(int[] flowerbed, int n) { int i = 0, count = 0; while (i < flowerbed.length) { if (flowerbed[i] == 0 && (i == 0 || flowerbed[i - 1] == 0) && (i == flowerbed.length - 1 || flowerbed[i + 1] == 0)) { flowerbed[i++] = 1; count++; } if(count>=n) return true; i++; } return false; } }
T :o(n) Space:O(1)
精简的写法
public class Solution { public boolean canPlaceFlowers(int[] flowerbed, int n) { int count =0; int len=flowerbed.length; for(int i=0;i<len;i++){ //判断当前是否为0,如果为0,且左右相邻均为0则可以种花,种花后修改花盆属性和种花数量 if(flowerbed[i]==0){ int right=(i==len-1)?0:flowerbed[i+1]; int left=(i==0)?0:flowerbed[i-1]; if(left==right&&right==0){ flowerbed[i]=1; count++; } } if(count>=n){ return true; } } return false; } }
时间: 2024-10-11 10:39:38