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  • 455. Assign Cookies 满足欲望 分配饼干

    [抄题]:

    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.

     [暴力解法]:

    时间分析:

    空间分析:

     [优化后]:

    时间分析:

    空间分析:

    [奇葩输出条件]:

    [奇葩corner case]:

    [思维问题]:

    读题错了,饼干能满足的>=人

    [一句话思路]:

    同向指针互相比较即可

    [输入量]:空: 正常情况:特大:特小:程序里处理到的特殊情况:异常情况(不合法不合理的输入):

    [画图]:

    [一刷]:

    [二刷]:

    [三刷]:

    [四刷]:

    [五刷]:

      [五分钟肉眼debug的结果]:

    [总结]:

    分配问题,用同向指针即可

    [复杂度]:Time complexity: O(n) Space complexity: O(1)

    [英文数据结构或算法,为什么不用别的数据结构或算法]:

    省空间

    [关键模板化代码]:

    [其他解法]:

    [Follow Up]:

    [LC给出的题目变变变]:

     [代码风格] :

    class Solution {
        public int findContentChildren(int[] g, int[] s) {
            //ini:sort
            int i = 0;
            Arrays.sort(g);
            Arrays.sort(s);
            
            //for loop
            for (int j = 0; j < s.length && i < g.length; j++) {
                if (s[j] >= g[i]) i++;
            }
            
            //return
            return i;
        }
    }
    View Code
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  • 原文地址:https://www.cnblogs.com/immiao0319/p/8973768.html
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