Leetcode-455Assign Cookies

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 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.

题解：

分糖果，给出一些孩子对糖果的需求，存入g；糖果的大小，存入s；求糖果所能满足的孩子需求的最多的数量。
考虑贪心的思想（选取当前最优的策略的算法），以Input: [3,1,2], [4,1]为例：孩子对糖果的需求分别为1，2，3；糖果大小为1，1；

1. 分别对g和s中的元素进行排序，Input变为 [1,2,3], [1,4]；
2. 依次将g, s中的元素进行对比，让最小的糖果尽可能的满足需求小的孩子，这样当孩子或者糖果中的一个比较完成后，输出满足分配条件的孩子数量。

My Solution（C/C++完整实现）：

#include <cstdio>
#include <iostream>
#include <vector>
#include <algorithm>

using namespace std;

class Solution {
public:
    int findContentChildren(vector<int> &g, vector<int> &s) {  //需求因子g, 糖果大小s
        sort(g.begin(), g.end());
        sort(s.begin(), s.end());
        int child = 0;
        int cookie = 0;
        while (child < g.size() && cookie < s.size()) {
            if (g[child] <= s[cookie]) {
                child += 1;
            }
            cookie += 1;
        }
        return child;
    }
};

int main() {
    Solution s;
    vector<int> children;
    vector<int> cookie;
    children.push_back(1);
    children.push_back(2);
    children.push_back(3);
    cookie.push_back(1);
    cookie.push_back(1);
    printf("%d\n", s.findContentChildren(children, cookie));
    return 0;
}

结果：

1

My Solution（Python）:

class Solution:
    def findContentChildren(self, g, s):
        """
        :type g: List[int]
        :type s: List[int]
        :rtype: int
        """
        g.sort()
        s.sort()
        child = cookie = result = 0
        while cookie < len(s) and child < len(g):
            if g[child] <= s[cookie]:
                result += 1
                child += 1
            cookie += 1
        return result

Reference:

class Solution:
    def findContentChildren(self, g, s):
        """
        :type g: List[int]
        :type s: List[int]
        :rtype: int
        """
        res = 0
        heapq.heapify(g)
        s.sort()
        for num in s:
            if not g:
                break
            elif g[0] <= num:
                res += 1
                heapq.heappop(g)
        return res