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LeetCode 455. Assign Cookies

2017-01-18 17:32 465 查看

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.

分析：先将g数组和s数组从小到大排序～ i指针遍历g数组，j指针遍历s数组，如果当前g[i] <= s[j]，也就是当前糖果j能够分给当前小朋友i，那就分配，并且将i指针指向下一个小朋友，cnt同时也要累加一个～如果当前糖果不能分配给当前小朋友，说明糖果j不能分给任何人了，因为没有比小朋友i需要的还要少的小朋友了（i后面的数都比i大）。所以无论是否分配，都把j向后移动一次～这样能保证需求量小的所有小朋友都分配得到能够分配的糖果，此时的cnt也是所求的贪心最大值～避免了大材小用（大糖果分配给需求量小的小朋友）的情况～

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

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