【贪心】C_009 分发饼干（暴力 | 双指针）

一、题目描述

Assume you are an awesome parent and want to give your children some cookies. But, you should give each child at most one cookie. Each child i has a greed factor gi, which is the minimum size of a cookie that the child will be content with; and each cookie j has a size sj. If sj >= gi, we can assign the cookie j to the child i, and the child i will be content. Your goal is to maximize the number of your content children and output the maximum number.

Note:

• You may assume the greed factor is always positive.
• You cannot assign more than one cookie to one child.

Input: [1,2], [1,2,3]
Output: 2
Explanation: You have 2 children and 3 cookies. The greed factors of 2 children are 1, 2. 
You have 3 cookies and their sizes are big enough to gratify all of the children, 
You need to output 2.

二、题解

方法一：排序

• 显然最优策略是每次尝试将尺寸 size 最小的饼干先发出去。
• 注意，发出去的饼干是不能在分配的了，所以这里用了一个 boolean[] vis 数组记录已分发的饼干。

public int findContentChildren(int[] g, int[] s) {
    Arrays.sort(g);
    Arrays.sort(s);
    int count = 0;
    boolean[] vis = new boolean[s.length];
    
    for (int i = 0; i < g.length; i++)
    for (int j = 0; j < s.length; j++) {
        if (s[j] >= g[i] && !vis[j]) {
            count++;
            vis[j] = true;
            break;
        }
    }
    return count;
}

复杂度分析

• 时间复杂度：$O(n\times m)$，这样做很容易想到，但是效率略低。
• 空间复杂度：$O(m)$，

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方法二：双指针

• 排完序之后，如果 s[cookie] 满足不了 g[child]，cookie 自然也就需要向后移动，寻找更大的饼干了。
• 如果 s[cookie] 可满足 g[child]，那么尝试满足下一个孩子即可。

public int findContentChildren(int[] g, int[] s) {
    Arrays.sort(g);
    Arrays.sort(s);
    int child = 0, cookie = 0;

    while (child < g.length && cookie < s.length) {
        if (g[child] <= s[cookie]) {
            child++;
        }
        cookie++;
    }    
    return child;
}

复杂度分析

• 时间复杂度：$O(nlogn)$，n 为孩子与饼干数量的最大者.
• 空间复杂度：$O(1)$，