题目描述:

Given an unsorted array of integers, find the length of the longest consecutive elements sequence.

For example,

Given [100, 4, 200, 1, 3, 2],

The longest consecutive elements sequence is [1, 2, 3, 4]. Return its length: 4.

Your algorithm should run in O(n) complexity.

最好想到的思路是排序,如果排序,最快也要O(NlogN),不符合要求。

以空间换时间:

代码:

  1. int longestConsecutive(vector<int>& nums)
  2. {
  3. map<int,int> map1;
  4. int len = 0;
  5. int maxlen = len;
  6. for(int i = 0;i<nums.size();i++)
  7. {
  8. //在nums中出现,则不为0,否则为0
  9. map1[nums[i]]++;
  10. }
  11. int min1 = nums[0];
  12. int min_up = min1;
  13. int min_down = min1-1;
  14. int i = 0;
  15. while(i<nums.size())
  16. {
  17. if(map1[min_up]!=0)
  18. {
  19. len++;
  20. map1[min_up] = 0;
  21. min_up++;
  22. }
  23. if(map1[min_down]!=0)
  24. {
  25. len++;
  26. map1[min_down] = 0;
  27. min_down--;
  28. }
  29. if(map1[min_down]==0 && map1[min_up]==0)
  30. {
  31. if(maxlen<len)
  32. maxlen = len;
  33. cout<<maxlen<<endl;
  34. len = 0;
  35. while(map1[nums[i++]]==0&&i<nums.size());
  36. if(i<nums.size())
  37. {
  38. min1 = nums[i-1];
  39. min_up = min1;
  40. min_down = min1-1;
  41. }
  42. else break;
  43. }
  44. }
  45. return maxlen;
  46. }

时间复杂度:O(N),同时引入了map,空间复杂度O(N).

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