采用贪心算法解题的简单例子

题目来源:

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题目描述:

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: 1Explanation: 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: 2Explanation: 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.

我的解决方案:

class Solution {public:    int GetLocation(vector<int>& nums,int low,int high)    {        int tmp=nums[low];        while(low<high)        {          while(low<high && tmp<nums[high]) high--;            nums[low]=nums[high];          while(low<high && tmp>=nums[low]) low++;            nums[high]=nums[low];         }        nums[low]=tmp;        return low;    }        void quick_sort(vector<int>& nums,int low,int high)    {        if(low<high)        {            int Loc=GetLocation(nums,low,high);            quick_sort(nums,low,Loc-1);            quick_sort(nums,Loc+1,high);        }    }        int findContentChildren(vector<int>& g, vector<int>& s) {        if(g.size()==0 || s.size()==0)          return 0;        int Ret=0;        quick_sort(g,0,g.size()-1);        quick_sort(s,0,s.size()-1);        for(int i=0,j=0;i<s.size() && j<g.size();++i)        {            if(s[i]>=g[j])            {                Ret++;                j++;            }        }        return Ret;    }};

思考:

cookie和child都不可拆分,所以如果两个序列都排好序,然后对孩子序列进行遍历,每个孩子拿走能满足他的最小的一块饼干,并标记拿走的位置,那么这个位置之前的饼干肯定不能满足需求更高的孩子,一直循环遍历直到孩子序列遍历完成或者拿走最靠后的一块饼干,就能得到最优的分配办法.这种求解思路其实就是贪心算法,只考虑当前情况下的最优解决方法,完成当前最优的求解之后继续向下求解,直到边界情况,合起来就是全局的解.代码实现上,自己重写了快速排序的函数来给这两个序列排序,实际也可以使用STL的sort方法来排序,看起来会简洁很多:

援引自:Yular

class Solution {public:    int findContentChildren(vector<int>& g, vector<int>& s) {        if(g.size() == 0 || s.size() == 0)            return 0;        int ans = 0, gid = 0;        sort(g.begin(), g.end());        sort(s.begin(), s.end());        for(int i = 0; i < s.size() && gid < g.size(); ++ i){            if(s[i] >= g[gid]){                ++ ans, ++ gid;            }        }        return ans;    }};