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.
You may assume the greed factor is always positive.
You cannot assign more than one cookie to one child.
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.
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.
先对两个数组按从大到小进行排序,判断饼干是否能让小孩开心,可以就记录一个,数组往后移动。如果不能则跳过这个孩子,对下一个进行比较。
class Solution {
public int findContentChildren(int[] g, int[] s) {
int gi = 0;
int si = 0;
int result = 0;
Arrays.sort(g);
Arrays.sort(s);
while(gi < g.length && si < s.length) {
if (s[si] >= g[gi]) {
result++;
gi++;
si++;
}else {
si++;
}
}
return result;
}
}class Solution {
public:
int findContentChildren(vector<int>& g, vector<int>& s) {
sort(g.begin(), g.end());
sort(s.begin(), s.end());
int result = 0;
int gi = 0;
int si = 0;
while (si < s.size() && gi < g.size()) {
if (s[si] >= g[gi]) {
si++;
gi++;
result++;
}else {
si++;
}
}
return result;
}
};