【leetcode】455. Assign Cookies
2017-04-15 23:38
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题目:
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.
思路:利用贪心算法,先将两个数组排序,之后让需求最小的孩子和供给最小的饼干比较,若孩子较小,则把饼干分配给孩子否则比较下一个饼干,以此类推。
代码:
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
class Solution {
public:
int findContentChildren(vector<int>& g, vector<int>& s) {
if (g.empty() || s.empty()) return 0;
sort(g.begin(),g.end());
sort(s.begin(), s.end());
int res = 0;
int j = 0;
int i = 0;
while (i < g.size()&& j < s.size()) {
if (g[i] <= s[j]) {
i++;
j++;
res++;
}
else {
j++;
}
}
return res;
}
};
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.
思路:利用贪心算法,先将两个数组排序,之后让需求最小的孩子和供给最小的饼干比较,若孩子较小,则把饼干分配给孩子否则比较下一个饼干,以此类推。
代码:
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
class Solution {
public:
int findContentChildren(vector<int>& g, vector<int>& s) {
if (g.empty() || s.empty()) return 0;
sort(g.begin(),g.end());
sort(s.begin(), s.end());
int res = 0;
int j = 0;
int i = 0;
while (i < g.size()&& j < s.size()) {
if (g[i] <= s[j]) {
i++;
j++;
res++;
}
else {
j++;
}
}
return res;
}
};
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