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： 两个无序数组，数组g是孩子们 ，数组s是饼干 。 每个蛋糕有自己的尺寸 。 每个孩子对蛋糕的尺寸有个最小要求，要让孩子接受这个饼干，那么尺寸必须达标。 任务就是如何分配这些蛋糕，使得尽量多的孩子接受你的饼干。返回能接受的孩子数量
思路： 首先对两个数组进行排序,然后按照优先满足贪婪度较低孩子的原则,若饼干大小能满足,则将它分配给该孩子,若不能,则看比它大一点的饼干能不能满足,而之前的饼干必然满足不了其他所有人的欲望.
代码如下：
代码1 C++ #include <iostream> #include <vector> #include<algorithm> using namespace std; class Solution { public: //g[] children s[] cookies //首先对两个数组进行排序,然后按照优先满足贪婪度较低孩子的原则, //若饼干大小能满足,则将它分配给该孩子, //若不能,则看比它大一点的饼干能不能满足,而之前的饼干必然满足不了其他所有人的欲望. //循环套循环 明显效率低 int findContentChildren(vector<int>& g, vector<int>& s) { int maxNum=0; sort(g.begin(), g.end()); sort(s.begin(), s.end()); int j=0; for (int i=0; i<g.size(); i++){ while(j < s.size()){ if(g[i]<=s[j]){ j++; maxNum++; break; } j++; } } return maxNum; } }; 代码2 //单循环 效率提高明显 int findContentChildren1(vector<int>& g, vector<int>& s) { int maxNum = 0; sort(g.begin(), g.end());//将两个数组排序 sort(s.begin(), s.end()); int i = 0; int j = 0; while(i<g.size() && j<s.size()){ if(g[i] <= s[j]){//不满足条件 maxNum+1 两个数组下标+1 maxNum++; i++; j++; } else{ j++;//不满足条件 s下标+1 继续和原g[i]比较 } } return maxNum; } 代码3 //代码2写出来后发现maxNum完全可以用i来取代 int findContentChildren2(vector<int>& g, vector<int>& s) { sort(g.begin(), g.end()); sort(s.begin(), s.end()); int i = 0; int j = 0; while(i<g.size() && j<s.size()){ if(g[i] <= s[j]){ i++; j++; } else{ j++; } } return i; } 代码4 int findContentChildren3(vector<int>& g, vector<int>& s) { //和上面方法差不多 int maxNum = 0; sort(g.begin(), g.end()); sort(s.begin(), s.end()); for(int Size:s) { if(Size >= g[maxNum]) { ++maxNum; if(maxNum == g.size()) break; } } return maxNum; } 代码5 //从数组后面遍历 int findContentChildren4(vector<int>& g, vector<int>& s) { int maxNum = 0; sort(g.begin(),g.end()); sort(s.begin(),s.end()); int i = g.size()-1; int j = s.size()-1; while(j >= 0){ while(i >= 0 && g[i]> s[j]) //找到符合条件的孩子下标 i--; if(i >= 0) //若符合条件的孩子下标在数组内 maxNum+1 maxNum++; j--; i--; } return maxNum; }注意：
sort排序头文件 algorithm