Flower problem
1. Title description
Suppose you have a very long flower bed, and one part of the plot is planted with flowers, but the other part is not. However, flowers cannot be planted on adjacent plots. They will compete for water and both will die.
Given a flower bed (represented as an array containing 0 and 1, where 0 means no flowers are planted, and 1 means flowers are planted), and a number n. Can n flowers be planted without breaking the planting rules? It returns True if it can, and False if it can't.
( LeetCode )
示例 1:
输入: flowerbed = [1,0,0,0,1], n = 1
输出: True
示例 2:
输入: flowerbed = [1,0,0,0,1], n = 2
输出: False
注意:
数组内已种好的花不会违反种植规则。
输入的数组长度范围为 [1, 20000]。
n 是非负整数，且不会超过输入数组的大小。
Two, the solution
1. Violence law (three numbers are considered together)
class Solution {
public boolean canPlaceFlowers(int[] flowerbed, int n) {
//处理长度小于3的情况
if(flowerbed.length==1){
return flowerbed[0]==0?(n<=1):(n==0);
}
if(flowerbed.length==2){
return flowerbed[0]==0&&flowerbed[1]==0?n<=1:n==0;
}
int count=0;
//左端需特殊考虑
if(flowerbed[0]==0&&flowerbed[1]==0){
flowerbed[0]=1;
count++;
}
for (int i = 1; i < flowerbed.length-1; i++) {
//将数组遍历三个三个数一起考虑
if(flowerbed[i-1]==0&&flowerbed[i]==0&&flowerbed[i+1]==0){
flowerbed[i]=1;
count++;
}
}
//右端特殊考虑
if(flowerbed[flowerbed.length-2]==0&&flowerbed[flowerbed.length-1]==0){
count++;
}
return count>=n;
}
}
2. Violence method (the two numbers are considered together, the special situation at the left end can be avoided, the method of commenting on the boss)
class Solution {
public boolean canPlaceFlowers(int[] flowerbed, int n) {
for (int i = 0; i < flowerbed.length; i += 2) {
if (flowerbed[i] == 0) {
//如果当前是最后一个数或者是当前数的下一个数是0
//说明当前可以种花，将n--（要种n朵，如果减到负值就说明不可以）
if (i == flowerbed.length - 1 || flowerbed[i + 1] == 0) {
n--;
} else {
//否则再跳一个数(本来就跳两个数，下一个数是1肯定也要跳过)
i++;
}
}
}
return n <= 0;
}
}
3. Greedy algorithm (official solution)
class Solution {
public boolean canPlaceFlowers(int[] flowerbed, int n) {
int count = 0;
int m = flowerbed.length;
//记录前一个为1的数组下标
int prev = -1;
for (int i = 0; i < m; i++) {
//如果当前为1进行判断
if (flowerbed[i] == 1) {
if (prev < 0) {
//当前下标前面的数没有1，count+=i / 2
//比如:[0,0,0,1] ,i=3,count+=1
count += i / 2;
} else {
//当前下标前面的数有1，count += (i - prev - 2) / 2;
//就是获取两个1之间最多放1的数
//比如：[1,0,0,0,1],pre=0,i=4,count+=1
count += (i - prev - 2) / 2;
}
prev = i;
}
}
if (prev < 0) {
//特殊情况：数组都为0
count += (m + 1) / 2;
} else {
//正常情况：最后边界有可能没统计进去
//如：[1,0,0,0,1,0,0](没统计最后一个数)
//此时：count=1,pre=4,m=7。count+=1
count += (m - prev - 1) / 2;
}
return count >= n;
}
}