[LeetCode-455]Assign Cookies

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本文介绍了一种解决分配饼干问题的方法，目标是最大化满足孩子的数量。通过排序和双指针技巧，实现高效匹配饼干与孩子的贪婪程度。

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.

Example 1:
Input: [1,2,3], [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:
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.

思路：首先将两个数组从小到大排序，设s数组中的最大元素为MaxS，然后在贪婪数组(g数组)中找到最后一个小于MaxS的元素的下标，记为x。

定义两个指针endg,ends分别指向g[x]和s[s.length - 1]，比较g[endg]与s[ends]的大小：
若g[endg] <= s[ends]，则结果 + 1，endg，ends分别前移一个单位；
若g[endg] > s[ends]，则当前孩子不能满足，则endg向前移动一个单位；
条件是 endg >= 0 && ends >=0

public class AssignCookies455 {
    public static int findContentChildren(int[] g, int[] s) {
        if (g == null || s == null || g.length == 0 || s.length == 0)
            return 0;

        Arrays.sort(g);
        Arrays.sort(s);
        int glen = g.length, slen = s.length;
        int maxS = s[slen - 1];
        int begin = 0, end = glen;
        while (begin < end) {
            int mid = begin + ((end - begin) >> 1);
            if (g[mid] > maxS + 1) {
                end = mid;
            } else {
                begin = mid + 1;
            }
        }

        int ends = slen - 1, endg = begin - 1, res = 0;
        while (ends >= 0 && endg >= 0) {
            if (g[endg] <= s[ends]) {
                res++;
                ends--;
            }

            endg--;
        }

        return res;
    }
}