05 August 2008
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:
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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:
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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.
1） 方法一，考虑以下情况
在以上的解法中，只在每次分配时饼干时选择一种看起来是当前最优的分配方法，但还不知道这种局部最优的分配方法最后是否能得到全局最优解。使用反证法进行证明（贪心法通常由反证法或数学归纳法证明），即假设存在一种比我们使用的贪心策略更优的最优策略。如果不存在这种最优策略，表示贪心策略就是最优策略，得到的解也就是全局最优解。
贪心法的正确性证明：假设在某次选择中，贪心策略选择给当前满足度最小的孩子分配第 m 个饼干，第 m 个饼干为可以满足该孩子的最小饼干。假设存在一种最优策略，可以给该孩子分配第 n 个饼干，并且 m < n。可以发现，经过这一轮分配，贪心策略分配后剩下的饼干一定有一个比最优策略来得大。因此在后续的分配中，贪心策略一定能满足更多的孩子。也就是说不存在比贪心策略更优的策略，即贪心策略就是最优策略。
排序 + Greedy
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class Solution {
public int findContentChildren(int[] g, int[] s) {
Arrays.sort(g);
Arrays.sort(s);
int g_i = 0;
int s_i = 0;
int result = 0;
while (g_i < g.length && s_i < s.length) {
if (g[g_i] <= s[s_i]) {
result++;
g_i++; // 满足了一个孩子
}
s_i++;
}
return result; // 这里返回g_i也行，用不着result
}
}