【leetcode系列】26. 刪除排序數組中的重複項

 

題目描述

給定一個排序數組，你須要在原地刪除重複出現的元素，使得每一個元素只出現一次，返回移除後數組的新長度。

 

不要使用額外的數組空間，你必須在原地修改輸入數組並在使用 O(1) 額外空間的條件下完成。

 

示例 1:

 

給定數組 nums = [1,1,2], 

 

函數應該返回新的長度 2, 而且原數組 nums 的前兩個元素被修改成 1, 2。 

 

你不須要考慮數組中超出新長度後面的元素。

示例 2:

 

給定 nums = [0,0,1,1,1,2,2,3,3,4],

 

函數應該返回新的長度 5, 而且原數組 nums 的前五個元素被修改成 0, 1, 2, 3, 4。

 

你不須要考慮數組中超出新長度後面的元素。

說明:

 

爲何返回數值是整數，但輸出的答案是數組呢?

 

請注意，輸入數組是以「引用」方式傳遞的，這意味着在函數裏修改輸入數組對於調用者是可見的。

 

你能夠想象內部操做以下:

 

1. // nums is passed in by reference. (i.e., without making a copy)

2. int len = removeDuplicates(nums);

3.  

4. // any modification to nums in your function would be known by the caller.

5. // using the length returned by your function, it prints the first len elements.

6. for (int i = 0; i < len; i++) {

7. print(nums[i]);

8. }

思路

使用快慢指針來記錄遍歷的座標。

• 開始時這兩個指針都指向第一個數字

• 若是兩個指針指的數字相同，則快指針向前走一步

• 若是不一樣，則兩個指針都向前走一步

• 當快指針走完整個數組後，慢指針當前的座標加1就是數組中不一樣數字的個數

關鍵點解析

• 雙指針

這道題若是不要求，O(n)的時間複雜度， O(1)的空間複雜度的話，會很簡單。 可是這道題是要求的，這種題的思路通常都是採用雙指針

• 若是是數據是無序的，就不能夠用這種方式了，從這裏也能夠看出排序在算法中的基礎性和重要性。

代碼

1. /*

2. * @lc app=leetcode id=26 lang=javascript

3. *

4. * [26] Remove Duplicates from Sorted Array

5. *

6. * https://leetcode.com/problems/remove-duplicates-from-sorted-array/description/

7. *

8. * algorithms

9. * Easy (39.76%)

10. * Total Accepted: 539.7K

11. * Total Submissions: 1.4M

12. * Testcase Example: '[1,1,2]'

13. *

14. * Given a sorted array nums, remove the duplicates in-place such that each

15. * element appear only once and return the new length.

16. *

17. * Do not allocate extra space for another array, you must do this by modifying

18. * the input array in-place with O(1) extra memory.

19. *

20. * Example 1:

21. *

22. *

23. * Given nums = [1,1,2],

24. *

25. * Your function should return length = 2, with the first two elements of nums

26. * being 1 and 2 respectively.

27. *

28. * It doesn't matter what you leave beyond the returned length.

29. *

30. * Example 2:

31. *

32. *

33. * Given nums = [0,0,1,1,1,2,2,3,3,4],

34. *

35. * Your function should return length = 5, with the first five elements of nums

36. * being modified to 0, 1, 2, 3, and 4 respectively.

37. *

38. * It doesn't matter what values are set beyond the returned length.

39. *

40. *

41. * Clarification:

42. *

43. * Confused why the returned value is an integer but your answer is an array?

44. *

45. * Note that the input array is passed in by reference, which means

46. * modification to the input array will be known to the caller as well.

47. *

48. * Internally you can think of this:

49. *

50. *

51. * // nums is passed in by reference. (i.e., without making a copy)

52. * int len = removeDuplicates(nums);

53. *

54. * // any modification to nums in your function would be known by the caller.

55. * // using the length returned by your function, it prints the first len

56. * elements.

57. * for (int i = 0; i < len; i++) {

58. * print(nums[i]);

59. * }

60. *

61. */

62. /**

63. * @param {number[]} nums

64. * @return {number}

65. */

66. var removeDuplicates = function(nums) {

67. const size = nums.length;

68. let slowP = 0;

69. for (let fastP = 0; fastP < size; fastP++) {

70. if (nums[fastP] !== nums[slowP]) {

71. slowP++;

72. nums[slowP] = nums[fastP]

73. }

74. }

75. return slowP + 1;

76. };