[Leetcode] Intersection of Two Linked Lists

Write a program to find the node at which the intersection of two singly linked lists begins.

For example, the following two linked lists:

A:          a1 → a2
                   ↘
                     c1 → c2 → c3
                   ↗            
B:     b1 → b2 → b3

begin to intersect at node c1.

Notes:

• If the two linked lists have no intersection at all, return null.
• The linked lists must retain their original structure after the function returns.
• You may assume there are no cycles anywhere in the entire linked structure.
• Your code should preferably run in O(n) time and use only O(1) memory.

Analysis:

The idea is pretty straightforward: count the length of list A and B, and move the longer list few steps from its head so that two list are at the same position to the end of the list. Move both lists toward the end of the lists and whenever they met, the node is the first node the intersection begins.

     ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
        int l1 = 0;
        ListNode* p1 = headA;
        while(p1 != NULL) {
            l1++;
            p1 = p1->next;
        }
        
        int l2= 0;
        ListNode* p2 = headB;
        while(p2 != NULL) {
            l2++;
            p2 = p2->next;
        }
        
        p1 = headA;
        p2 = headB;
        if (l1 > l2) {
            for(int i=0; i<l1-l2; i++)
                p1 = p1->next;
        }
        else {
            for(int i=0; i<l2-l1; i++)
                p2 = p2->next;            
        }
        
        while(p1 != NULL && p2 != NULL && p1 != p2)
        {
            p1 = p1->next;
            p2 = p2->next;
        }
            
        return p1;
    }

[Leetcode] Find Minimum in Rotated Sorted Array II

Suppose a sorted array is rotated at some pivot unknown to you beforehand.

(i.e., 0 1 2 4 5 6 7 might become 4 5 6 7 0 1 2).

Find the minimum element.

The array may contain duplicates.

Analysis:

Similar to Find Minimum in Rotated Sorted Array, except that you need to take care of duplicate one.

int findMin(vector &num) {
        int b = 0;
        int e = num.size()-1;
        int minVal = num[0];
        while(b <= e)
        {
            int m = (b+e)/2;
            if (num[m] > num[b])
            {
                minVal = min(minVal, num[b]);
                b = m + 1;
            }
            else if (num[m] < num[e])
            {
                minVal = min(minVal, num[m]);
                e = m - 1;
            }
            else if (num[m] == num[e])
            {
                minVal = min(minVal, num[m]);
                e--;
            }
            else if (num[m] == num[b])
            {
                minVal = min(minVal, num[m]);
                b++;
            }
        }
        return minVal;        
    }