Consideration make three pointers , pre,current,next; initial pre as null use tmp to save current’s next node info change current’s next to link pre node(first is null) move pre pointer to next node move current pointer to next node soultion 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 /** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode(int x) { val = x; } * } */ class Solution { public ListNode reverseList(ListNode head) { ListNode pre = null; ListNode current = head; while (current != null){ ListNode next = current.next; current.next = pre; pre = current; current = next; } return pre; } }……

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The Hamming distance between two integers is the number of positions at which the corresponding bits are different. Given two integers x and y, calculate the Hamming distance. Note: 0 ≤ x, y < 231. Example: Input: x = 1, y = 4 Output: 2 Explanation: 1 (0 0 0 1) 4 (0 1 0 0) ↑ ↑ The above arrows point to positions where the corresponding bits are different. Consideration This problem is also have a relationship with ‘^’ , Think about it : 1(0001) and 4(0100) their Xor is 5(0101) . next we use & Operator to calculate the number of 1 , let the binary & 1 ,if result is 1 ,sum’s up , use » to move the postion. 5(0101) 0&1 = 0 pass 01&01 = 1 sum 010&001 = 0 pass 0101&0001 = 1 sum……

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