LeetCode-455. Assign Cookies
2018-03-29 19:14
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Description
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.
Solution 1(C++)
class Solution { public: int findContentChildren(vector<int>& g, vector<int>& s) { sort(g.begin(), g.end()); sort(s.begin(), s.end()); int num=0; for(int i=0, j=0; i<g.size() && j<s.size(); j++){ if(g[i]<s[j] || g[i]==s[j]){num++; i++;} } return num; } };
Solution 2(C++)
static int x = []() { std::ios::sync_with_stdio(false); cin.tie(NULL); return 0; }(); class Solution { int findge(int*nums, int v, int s, int e) const { while (s < e) { int m = s + ((e - s) >> 1), vv = nums[m]; if (vv >= v) { e = m; } else { s = m + 1; } } return s == e && nums[s] >= v ? s : INT_MAX; }; public: int findContentChildren(vector<int>& g, vector<int>& s) const { int ns = (int)s.size(); if (ns --) { int nc = (int)g.size(); if (nc) { int index = 0, count = 0, *ps = &s[0], *pg = &g[0], *pge = pg + nc; std::sort(g.begin(), g.end()); std::sort(s.begin(), s.end()); while (pg < pge) { index = findge(ps, *pg++, index, ns); if (index == INT_MAX) { return count; } ++index; ++count; } return count; } } return 0; }; };
Solution 3(C++)
class Solution { public: int findContentChildren(vector<int>& g, vector<int>& s) { priority_queue<int> gPq; priority_queue<int> sPq; for(int& gi : g) { gPq.push(gi); } for(int& si : s) { sPq.push(si); } int matched = 0; while(gPq.size() > 0 && sPq.size() > 0) { if(gPq.top() > sPq.top()) { gPq.pop(); continue; // cannot satisfy this kid } else { //(gPq.top() <= sPq.top()) ++matched; sPq.pop(); gPq.pop(); } } return matched; } };
算法分析
解法一是自己写的,速度还不错,解法二是最快的,与LeetCode-155. Min Stack 的解法二的第一句都一样。还是很有意思的。解法三用到了queue,目前还没刷到。所以不着急啦。程序分析
总结一个做这道题犯的错误,j错写成i,i也有可能误写成j。相关文章推荐
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