【贪心】B_013 种花问题（错误枚举 | 防御式编程 | 模拟）

一、题目描述

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.

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.

二、题解

方法一：错误枚举

遇到 1 就枚举前两个位置是否为 0，但是没有当前位置是 0，后序位置没有 1 的情况。 即这种情况：

[0,0,1,0,1]
1

public boolean canPlaceFlowers(int[] f, int n) {
    int count = 0, N = f.length;

    for (int i = 0; i < N; i++) {
        if (f[i] == 1)
            continue;
        for (int j = i + 2; j < N; j += 2) {
            if (f[j] == 0 && f[j - 1] == 0 && f[j + 1] == 0) {
                count++;
                f[j] = 1;
                break;
            }
        }
    }
    return count >= n;
}

防御式编程： 一种比较直观的做法就是在数组 f 的首尾各加 1 个前导 0，这样就不必考虑边界。

复杂度分析

• 时间复杂度：$O(n)$，
• 空间复杂度：$O(n)$，

---

方法二：贪心

• 遇到 0 就判断它的前后是否为 0。
  • 如果 f[i-1] 与 f[i+1] 都为 0 则可种。
  • 否则，跳过。
• 边界问题：
  • 若 i = 0，则默认 last 为 0.
  • 若 i = N-1，则默认 next = 0

public boolean canPlaceFlowers(int[] f, int n) {
    int count = 0, N = f.length;
    for (int i = 0; i < N; i++) {
        if (f[i] == 1)
            continue;
        int last = i == 0 ? 0 : f[i-1];
        int next = i == N-1 ? 0 : f[i+1];
        if (last == 0 && next == 0) {
            count++;
            f[i] = 1;
        } 
    }
    return count >= n;
}

复杂度分析

• 时间复杂度：$O(n)$，
• 空间复杂度：$O(1)$，