Leetcod-455_Assign Cookies (分饼干)-贪心算法-【C++】

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.

题目大意：
假设你是一个了不起的家长，想给你的孩子一些饼干。但是，你应该给每个孩子最多一个饼干。每个孩子我都有一个贪婪因子gi，这是孩子满意的一个cookie的最小大小；每个Cookie j有一个大小Sj。如果sj>=gi，我们可以将cookie j分配给子表i，而子表i将满足。您的目标是最大限度地增加内容子程序的数量，并输出最大数量的内容。

解题思路：
典型的贪心算法题，有两种常见的解题思路，
①先尝试满足最贪心的小朋友
②先尝试满足最不贪心的小朋友

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

using namespace std;

/// 455. Assign Cookies
/// https://leetcode.com/problems/assign-cookies/description/
/// 先尝试满足最贪心的小朋友
/// 时间复杂度: O(nlogn)
/// 空间复杂度: O(1)
class Solution {
public:
    int findContentChildren(vector<int>& g, vector<int>& s) {

        sort(g.begin(), g.end(), greater<int>());
        sort(s.begin(), s.end(), greater<int>());

        int gi = 0, si = 0;
        int res = 0;
        while(gi < g.size() && si < s.size()){
            if(s[si] >= g[gi]){
                res ++;
                si ++;
                gi ++;
            }
            else
                gi ++;
        }

        return res;
    }
};

int main() {

    int g1[] = {1, 2, 3};
    vector<int> gv1(g1, g1 + sizeof(g1)/sizeof(int));
    int s1[] = {1, 1};
    vector<int> sv1(s1, s1 + sizeof(s1)/sizeof(int));
    cout << Solution().findContentChildren(gv1, sv1) << endl;

    int g2[] = {1, 2};
    vector<int> gv2(g2, g2 + sizeof(g2)/sizeof(int));
    int s2[] = {1, 2, 3};
    vector<int> sv2(s2, s2 + sizeof(s2)/sizeof(int));
    cout << Solution().findContentChildren(gv2, sv2) << endl;

    return 0;
}

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

using namespace std;

/// 455. Assign Cookies
/// https://leetcode.com/problems/assign-cookies/description/
/// 先尝试满足最不贪心的小朋友
/// 时间复杂度: O(nlogn)
/// 空间复杂度: O(1)
class Solution {
public:
    int findContentChildren(vector<int>& g, vector<int>& s) {

        sort(g.begin(), g.end());
        sort(s.begin(), s.end());

        int gi = 0, si = 0;
        int res = 0;
        while(gi < g.size() && si < s.size()){
            if(s[si] >= g[gi]){
                res ++;
                si ++;
                gi ++;
            }
            else
                si ++;
        }

        return res;
    }
};

int main() {

    int g1[] = {1, 2, 3};
    vector<int> gv1(g1, g1 + sizeof(g1)/sizeof(int));
    int s1[] = {1, 1};
    vector<int> sv1(s1, s1 + sizeof(s1)/sizeof(int));
    cout << Solution().findContentChildren(gv1, sv1) << endl;

    int g2[] = {1, 2};
    vector<int> gv2(g2, g2 + sizeof(g2)/sizeof(int));
    int s2[] = {1, 2, 3};
    vector<int> sv2(s2, s2 + sizeof(s2)/sizeof(int));
    cout << Solution().findContentChildren(gv2, sv2) << endl;

    return 0;
}