LeetCode-455

原题链接

描述

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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.

示例

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Example 1:

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:

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.

Constraints:

• 1 <= g.length <= 3 * 104
• 0 <= s.length <= 3 * 104
• 1 <= g[i], s[j] <= 231 - 1

思路

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贪心。把 g 数组和 s 数组排序，然后从最小的饼干开始和最小的食欲进行比较，如果饼干份额大于等于食欲，那么记录满足人数 +1 并对下一块饼干和下一份食欲进行比较，否则用更大的饼干和当前人的食欲进行比较，直到一方为空。

解答

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class Solution {
public static void qSort(int[] arr, int low, int high) {
        if (low > high) return;
        int i = low, j = high, tmp = arr[low];

        while (i < j) {
            while (tmp <= arr[j] && i < j)
                j--;
            while (tmp >= arr[i] && i < j)
                i++;
            if (i < j) {
                int t = arr[i];
                arr[i] = arr[j];
                arr[j] = t;
            }
        }
        arr[low] = arr[i];
        arr[i] = tmp;
        qSort(arr, low, j - 1);
        qSort(arr, j + 1, high);
    }

    public static int findContentChildren(int[] g, int[] s) {
        if (g.length == 0 || s.length == 0) return 0;
        qSort(s, 0, s.length - 1);
        qSort(g, 0, g.length - 1);
        int cnt = 0;
        int i = 0, j = 0;
        while (i < s.length && j < g.length) {
            if (s[i] >= g[j]) {
                cnt++;
                j++;
            }
            i++;
        }
        return cnt;
    }
}