455. Assign Cookies

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 g[i], which is the minimum size of a cookie that the child will be content with; and each cookie j has a size s[j]. If s[j] >= g[i], 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.

Example 1:

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Input: g = [1,2,3], s = [1,1]
Output: 1
Explanation: You have 3 children and 2 cookies. The greed factors of 3 children are 1, 2, 3. 
And even though you have 2 cookies, since their size is both 1, you could only make the child whose greed factor is 1 content.
You need to output 1.

Example 2:

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Input: g = [1,2], s = [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.

题目大意

给定两个数组：g（每个元素表示孩子的贪心因子，即孩子满足的最小饼干尺寸）和s（每个元素表示饼干的尺寸）。每个孩子最多分配一块饼干，只有当饼干尺寸s[j] >= g[i]时，孩子i才会满足。目标是找到能让最多孩子满足的数量并返回。

例如：

  • 输入g = [1,2,3], s = [1,1]，输出1（仅贪心因子为 1 的孩子能得到尺寸为 1 的饼干）；
  • 输入g = [1,2], s = [1,2,3]，输出2（两个孩子的贪心需求都能被满足）。

解题思路

核心思路是贪心算法，通过 “小饼干优先满足小贪心” 的策略最大化满足的孩子数量，具体步骤如下：

  1. 排序预处理：
    • 对孩子的贪心因子数组g按升序排序（从小到大处理孩子的需求）；
    • 对饼干尺寸数组s按升序排序（从小到大使用饼干，避免大饼干浪费在小需求上）。
  2. 双指针匹配：
    • 用两个指针i（指向当前未满足的孩子，初始为 0）和j（指向当前未分配的饼干，初始为 0）；
    • 遍历饼干数组：
      • 若当前饼干s[j] >= g[i]：该饼干能满足当前孩子，i和j同时后移（孩子满足，饼干分配）；
      • 若当前饼干s[j] < g[i]：该饼干太小，无法满足当前及后续孩子（因数组已排序），仅j后移（跳过该饼干）；
    • 当i遍历完所有孩子或j遍历完所有饼干时，停止匹配。
  3. 结果返回：
    • 指针i的最终值即为满足的孩子数量（i从 0 开始，每满足一个孩子加 1）。

代码实现

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class Solution {
public:
    int findContentChildren(vector<int>& g, vector<int>& s) {
        // 对贪心因子和饼干尺寸排序
        sort (g.begin (), g.end ());
        sort (s.begin (), s.end ());
        int i = 0, j = 0; //i: 当前未满足的孩子索引，j: 当前未分配的饼干索引
        int childCount = g.size (), cookieCount = s.size ();
        while (i < childCount && j < cookieCount) {
            // 饼干能满足当前孩子，分配并处理下一个孩子和饼干
            if (s [j] >= g [i]) {
                i++;
                j++;
            }
            // 饼干太小，跳过当前饼干
            else {
                j++;
            }
        }
        return i; //i 的值即为满足的孩子数量
    }
};