455. Distribute cookies [double pointer ++i, ++j]

455. Distribute cookies
Suppose you are a great parent , Want to give your kids some cookies . however , Each child can only give one biscuit at most .

For every child i, All have an appetite value g[i], This is the smallest size of biscuit that can satisfy children's appetite ; And every cookie j, They all come in one size s[j] . If s[j] >= g[i], We can put this biscuit j Assign to children i , The child will be satisfied . Your goal is to meet as many children as possible , And output the maximum value .

Example 1:
Input : g = [1,2,3], s = [1,1]
Output : 1
explain :
You have three children and two biscuits ,3 The appetites of a child are ：1,2,3.
Although you have two biscuits , Because they are all of the same size 1, You can only make your appetite worth 1 The children of .
So you should output 1.

Example 2:
Input : g = [1,2], s = [1,2,3]
Output : 2
explain :
You have two children and three biscuits ,2 The appetites of a child are 1,2.
You have enough cookies and sizes to satisfy all children .
So you should output 2.

Tips ：
1 <= g.length <= 3 * 104
0 <= s.length <= 3 * 104
1 <= g[i], s[j] <= 231 - 1

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Ideas ：

Find the biggest appetite that cookies can satisfy g[i] < s[j], When traversing, the condition can only g[i] > s[j] :

Ergodic time , Take the appetite value as the benchmark , Traverse Biscuit size , The pointer j, With g[i] > s[j] Traverse the condition , After traversing the cookie size, it is over .

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Java Code ：

class Solution {
    
    public int findContentChildren(int[] g, int[] s) {
    
        Arrays.sort(g);
        Arrays.sort(s);
        int numOfChildren = g.length;
        int 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;
    }
}

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Js Code ：

var findContentChildren = function(g, s) {
    
    g.sort((a,b) => a-b);  // () Return results 
    s.sort((a,b) => a-b);
    let numOfChild = g.length, numOfCookies = s.length;
    let count=0;
    for(let i = 0, j = 0; i < numOfChild && j < numOfCookies; i++, j++) {
    
        while(j < numOfCookies && g[i] > s[j]) {
    
            j++; // Traverse j The words are equivalent to g[i] Set the point , scanning s[j]
        }
        if(j < numOfCookies){
    
            count++;
        }
    }
    return count;
};