Source: Link: https://leetcode-cn.com/problems/assign-cookies
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Suppose you are a great parent and want to give your children some cookies. However, each child can only give one cookie at most.
For each child i, there is an appetite value g[i], which is the smallest size of biscuit that can satisfy the appetite of the children; and each biscuit j has a size s[j]. If s[j] >= g[i], we can assign this cookie j to child i, and this child will be satisfied. Your goal is to satisfy as many children as possible and output this maximum value.
Example 1:
Input: g = [1,2,3], s = [1,1]
Output: 1
Explanation:
You have three children and two cookies. The appetite values of the three children are: 1, 2, 3.
Although you have two small biscuits, since their size is 1, you can only satisfy the child whose appetite is 1.
So you should output 1.
Example 2:
Input: g = [1,2], s = [1,2,3]
Output: 2
Explanation:
You have two children and three biscuits, and the appetite values of the two children are 1,2 respectively.
The number and size of cookies you have are enough to satisfy all children.
So you should output 2.
提示：
1 <= g.length <= 3 * 10^4
0 <= s.length <= 3 * 10^4
1 <= g[i], s[j] <= 2^31 - 1
My code double pointer
Here is to sort the two arrays first, the two pointers point to the beginning of the array, and compare them one by one
- g[i] <= s[j] means that the candy can be allocated to g[i], ret++;
- g[i]> s[j] means that the candy cannot be divided into g[i], [j++] points to the next candy
class Solution {
public int findContentChildren(int[] g, int[] s) {
//Arrays.sort(g);
quickSort(g, 0, g.length-1);
quickSort(s, 0, s.length-1);
//Arrays.sort(s);
int i=0, j=0;
int ret=0;
while(i<g.length && j<s.length){
if(g[i] <= s[j]){
ret++;
i++; j++;
}else{
j++;
}
}
return ret;
}
public void quickSort(int[] arr, int start, int end){
if(start >= end) return;
int p = partition(arr, start, end);
quickSort(arr, start, p);
quickSort(arr, p+1, end);
}
public int partition(int[]arr, int left, int right){
int key = arr[left];
while(left < right){
while(left < right && arr[right] >= key) right--;
arr[left] = arr[right];
while(left < right && arr[left] <= key) left ++;
arr[right] = arr[left];
}
arr[left] = key;
return left;
}
}
Great God's Code
This is the idea,
https://leetcode-cn.com/problems/assign-cookies/solution/fen-fa-bing-gan-by-leetcode-solution-50se/
class Solution {
public int findContentChildren(int[] g, int[] s) {
Arrays.sort(g);
Arrays.sort(s);
int numOfChildren = g.length, numOfCookies = s.length;
int count = 0;
for (int i = 0, j = 0; i < numOfChildren && j < numOfCookies; i++, j++) {
while (j < numOfCookies && g[i] > s[j]) {
j++;
}
if (j < numOfCookies) {
count++;
}
}
return count;
}
}
作者：LeetCode-Solution
链接：https://leetcode-cn.com/problems/assign-cookies/solution/fen-fa-bing-gan-by-leetcode-solution-50se/
来源：力扣（LeetCode）
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