leetcode540

Timeout speed pointer , But it timed out .. Purring

int singleNonDuplicate(int* nums, int numsSize){
    // Speed pointer , If the previous bit is equal to the current one , That handle fast Assign a value to slow, Move forward 
    int slow=0,fast=1;
    int ans=0;
    int len=numsSize;
    if(len==1)
    {
        return nums[0];
    }
    while(slow<len)
    {
        if(nums[slow]==nums[fast])
        {
            slow=fast;
        }
        else
        {
            ans=nums[slow];
            break;
        }
        slow++;
        fast+=2;// Take two steps at a time , Because the title says that every element will appear twice 
        // Find out that only 1 Time of , Can pass fast In even digits and fast Compare in odd digits 
    }
     return ans;
}

int singleNonDuplicate(int* nums, int numsSize){
  for(int i=0;i<numsSize-1;i+=2)
  {
      if(nums[i]!=nums[i+1])
      return nums[i];
  }
  return nums[numsSize-1];
}

int singleNonDuplicate(int* nums, int numsSize){
  int left=0;
  int right=numsSize;
  while(left<right)
  {
      int mid=left+(right-left)/2;
      if(mid%2==0)//mid For even when , The second is on the right 
      {
          if(mid+1<numsSize&&nums[mid]==nums[mid+1])
          {
              left=mid+1;
          }
          else{
              right=mid;
          }
          }
         
       else{// In an odd number of , The first one is on the left 
              if(mid-1>=0&&nums[mid-1]==nums[mid])
              {
                  left=mid+1;
              }
              else{
                  right=mid;
              }
          }
      
  }
  return nums[right];
}