Implement strStr().

Return the index of the first occurrence of needle in haystack, or -1 if needle is not part of haystack.

Example 1:

Input: haystack = "hello", needle = "ll"
Output: 2

Example 2:

Input: haystack = "aaaaa", needle = "bba"
Output: -1

Clarification:

What should we return when needle is an empty string? This is a great question to ask during an interview.

For the purpose of this problem, we will return 0 when needle is an empty string. This is consistent to C’s strstr() and Java’s indexOf().

public class Solution {
    public int strStr(String haystack, String needle) {  
        for (int i = 0;; i++) {  
            for (int j = 0;; j++) {  
                if (j == needle.length())  
                    return i;  
                if (i + j == haystack.length())  
                    return -1;  
                if (needle.charAt(j) != haystack.charAt(i + j))  
                    break;  
            }  
        }  
    }  
}
class Solution {
    public int strStr(String haystack, String needle) {
        if(needle.length() == 0 || needle == null || haystack.equals(needle)) return 0;
        if(haystack.length() < needle.length() || !haystack.contains(needle)) return -1;

        int i = 0;
        while(i<haystack.length()){
            
            int aindex = i;
            int j = 0;
            
            while(j<needle.length()){
                char a = haystack.charAt(aindex);
                char b = needle.charAt(j);
                if(a==b){
                    aindex++;
                    j++;
                }
                else{
                    break;
                }
                if(j >= needle.length())
                    return i;
            }
            i++;
        }
        return -1;
    }
}
/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
        ListNode node = new ListNode();
        ListNode root = node;
        while(l1!=null && l2!=null) {
            if(l1.val <= l2.val)
            {
                node.next = l1;
                l1 = l1.next;
            } else {
                node.next = l2;
                l2 = l2.next;
            }
            node = node.next;
        }
        if(l1 == null)
            node.next = l2;
        if(l2 == null)
            node.next = l1;
        
        return root.next;
    }
}