https://leetcode.com/problems/assign-cookies/數組
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.cookie
Note: You may assume the greed factor is always positive. You cannot assign more than one cookie to one child.測試
Example 1:code
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:排序
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.
題目叫「分配餅乾」,給定兩個數組,分別表示每一個小孩指望的餅乾尺寸,和每一個餅乾實際的尺寸。將餅乾分配給這些小孩,但分配的餅乾尺寸必須不小於小孩指望的餅乾尺寸。求出這些餅乾最多能夠知足幾個小孩。leetcode
解法:題目比較簡單清晰,既然是餅乾尺寸不小於指望尺寸,那麼咱們首先將兩個數組先排序。遍歷指望尺寸的數組和餅乾尺寸的數組,若是餅乾尺寸符合則兩個數組都向前進一,表示有一個餅乾知足了一個小孩;若是餅乾尺寸不符合,則餅乾數組向前進一,嘗試下一個餅乾。get
public int findContentChildren(int[] g, int[] s) { int ret = 0; Arrays.sort(g); Arrays.sort(s); int i = 0, j = 0; while (i < g.length && j < s.length) { if (g[i] <= s[j]) { ret++; i++; j++; }else if (g[i] > s[j]) { j++; } } return ret; }
測試用例:it
public static void main(String[] args) { Solution so = new Solution(); { int []g = {1, 2, 3}; int []s = {1, 1}; assert(so.findContentChildren(g, s) == 1); } { int []g = {1, 2}; int []s = {1, 2, 3}; assert(so.findContentChildren(g, s) == 2); } { int []g = {}; int []s = {}; assert(so.findContentChildren(g, s) == 0); } }