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.
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.
解析
题目中要求更多的分配cookies，即“贪心”。实现的思路比较简单，根据题目中的逻辑进行实现即可，即从左至右遍历s，如果比g大，即满足该child。
实现（C语言）
int comp(const void* a, const void* b) {
return *(int*)a > *(int*)b;
}
int findContentChildren(int* g, int gSize, int* s, int sSize) {
qsort(g, gSize, sizeof(int), comp);
qsort(s, sSize, sizeof(int), comp);
int count = 0, j = 0;
for (int i = 0; i < sSize; ++i) {
if (s[i] >= g[j]) {
++count;
++j;
if (j >= gSize)
break;
}
}
return count;
}
实现中使用到了qsort库函数，即将数组进行从大到小排序，先从大到小进行满足，可以更多的满足child，这个是显然的。