/*** Definition for singly-linked list.* public class ListNode {*     int val;*     ListNode next;*     ListNode(int x) {*         val = x;*         next = null;      *     }* }*/
public class Solution {/*** @param l1: the first list* @param l2: the second list* @return: the sum list of l1 and l2 */public ListNode addLists(ListNode l1, ListNode l2) {// write your code here
        ListNode head= new ListNode(0);ListNode current = head;int quotients=0;while(l1!=null && l2!=null){int sum = quotients + l1.val + l2.val;quotients = sum/10;// l1.val = sum%10;current.next = new ListNode(sum%10);current = current.next;l1 = l1.next;l2 = l2.next;}while(l1!=null){int sum = quotients + l1.val;quotients = sum/10;current.next = new ListNode(sum%10);current = current.next;l1 = l1.next;}while(l2!=null){int sum = quotients + l2.val;quotients = sum/10;current.next = new ListNode(sum%10);current = current.next;l2 = l2.next;}if(quotients==1){current.next = new ListNode(1);current = current.next;}return head.next;}
}

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = Noneclass Solution:# @param l1: the first list# @param l2: the second list# @return: the sum list of l1 and l2 def addLists(self, l1, l2):# write your code herehead = ListNode(0)current = head;sum = 0while l1!=None or l2!=None:if l1!=None:sum+=l1.vall1=l1.nextif l2!=None:sum+=l2.vall2=l2.nextcurrent.next = ListNode(sum%10)current = current.nextsum/=10if sum==1:current.next = ListNode(1)current = current.nextreturn head.next

给出数组[4, 5, 1, 2, 3]， 返回 3

给出数组[7, 9, 4, 5]，返回 5

时间复杂度为O(n)

public class Solution {/*** @param nums: a list of integers* @return: find a  majority number*/public int majorityNumber(ArrayList<Integer> nums) {// write your codeint numsLen = nums.size();HashMap hmap = new HashMap();int majority = 0;for(int i = 0;i<numsLen;i++){int num = nums.get(i);if(hmap.containsKey(num)==false){hmap.put(num,1);}else{hmap.put(num,(Integer)hmap.get(num)+1);if((Integer)hmap.get(num)>numsLen/2){majority= num;break;}}}return majority;}
}

public class Solution {/*** @param nums: a list of integers* @return: find a  majority number*/public int majorityNumber(ArrayList<Integer> nums) {// write your codeint numsLen = nums.size();HashMap hmap = new HashMap();int majority = 0;for(int i = 0;i<numsLen;i++){int num = nums.get(i);if(hmap.containsKey(num)==false)hmap.put(num,1);elsehmap.put(num,(Integer)hmap.get(num)+1);if((Integer)hmap.get(num)>numsLen/2){majority= num;break;}}return majority;}
}

class Solution:"""@param nums: A list of integers@return: The majority number"""def majorityNumber(self, nums):# write your code herenumsLen = len(nums)d={}for num in nums:if num not in d:d[num] = 1else:d[num] +=1if d[num]>numsLen/2:return num

public class Solution {/*** @param nums: a list of integers* @return: find a  majority number*/public int majorityNumber(ArrayList<Integer> nums) {// write your codeint numsLen = nums.size();int mid = numsLen/2;int left = 0;int right = numsLen - 1;return findKth(nums,mid,left,right);}int findKth(ArrayList<Integer> nums,int mid,int left,int right){if(mid==left && mid == right)return nums.get(mid);if(left>=right)return -1;int i=left;int j=right;int pivot=nums.get(left);while(i<j){while(i<j && nums.get(j)>pivot) j--;if(i<j){nums.set(i,nums.get(j));i++;}while(i<j && nums.get(i)<pivot) i++;if(i<j){nums.set(j,nums.get(i));j--;}}nums.set(i,pivot);if(i==mid) return nums.get(mid);else if(i>mid)return findKth(nums,mid,left,i-1);else return findKth(nums,mid,i+1,right);}
}

public class Solution {/*** @param nums: a list of integers* @return: find a  majority number*/public int majorityNumber(ArrayList<Integer> nums) {// write your codeint numsLen = nums.size();if(numsLen==1) return nums.get(0);int times=0;int maj = 0;for(int i=0;i<numsLen;i++){if(times==0){maj = nums.get(i);times= 1;}else{if(maj==nums.get(i))times++;elsetimes--;}}return maj;}
}

class Solution:"""@param nums: A list of integers@return: The majority number"""def majorityNumber(self, nums):# write your code herenumsLen = len(nums)if numsLen == 1 :return nums[0]times = 0maj = 0for num in nums:if times==0:times=1maj = numelif maj==num:times +=1else:times -=1return maj

public class Solution {/*** @param nums: a list of integers* @return: find a  majority number*/public int majorityNumber(ArrayList<Integer> nums) {// write your codeint first = Integer.MIN_VALUE; // the number with the top votesint firstVote = 0; // the top votesint second = Integer.MIN_VALUE; // the number with the second top votesint secondVote = 0; // the second top votesint numsLen = nums.size();if(numsLen==1)return nums.get(0);for(int i=0;i<numsLen;i++){if(firstVote > 0 && nums.get(i) == first) {firstVote++;} else if(secondVote > 0 && nums.get(i) == second) {secondVote++;if(secondVote > firstVote) {int t = first;first = second;second = t;t = firstVote;firstVote = secondVote;secondVote = t;}} else if(firstVote == 0) {first = nums.get(i);firstVote++;} else if(secondVote == 0) {second = nums.get(i);secondVote++;}  else {firstVote--;secondVote--;}}// confirm if the number with top 2 votes occurs more than 3/n times// int firstCount = 0;// int secondCount = 0;// for(int i=0;i<numsLen;i++) {// if(firstVote > 0 && nums.get(i) == first) firstCount++;// if(secondVote > 0 && nums.get(i) == second) secondCount++;// }// if(firstCount > numsLen / 3) return first;// else if(secondCount > numsLen / 3) return second; // else return 0;return first;}
}