刷題日記（13）Remove Duplicates from Sorted Array
3 min readJan 5, 2021
Given a sorted array nums, remove the duplicates in-placesuch that each element appears only once and returns the new length.
Do not allocate extra space for another array, you must do this by modifying the input array in-place with O(1) extra memory.
Clarification:
Confused why the returned value is an integer but your answer is an array?
Note that the input array is passed in by reference, which means a modification to the input array will be known to the caller as well.
Internally you can think of this:
// nums is passed in by reference. (i.e., without making a copy)
int len = removeDuplicates(nums);// any modification to nums in your function would be known by the caller.
// using the length returned by your function, it prints the first len elements.
for (int i = 0; i < len; i++) {
print(nums[i]);
}
Example 1:
Input: nums = [1,1,2]
Output: 2, nums = [1,2]
Explanation: Your function should return length = 2, with the first two elements of nums being 1 and 2 respectively. It doesn't matter what you leave beyond the returned length.Example 2:
Input: nums = [0,0,1,1,1,2,2,3,3,4]
Output: 5, nums = [0,1,2,3,4]
Explanation: Your function should return length = 5, with the first five elements of nums being modified to 0, 1, 2, 3, and 4 respectively. It doesn't matter what values are set beyond the returned length.Constraints:
0 <= nums.length <= 3 * 104-104 <= nums[i] <= 104numsis sorted in ascending order.
解題思路：
- “in-place” 所以用雙指針，而且條件說list已經排好了
- 一根指針讀，一根指針寫
- 當快指針的值不等於慢指針時，慢指針加一格寫入快指針的值，之後快指針繼續走
- 注意不要不小心用到三指針，for loop裡面的i就可以當一個指針了
- 最後回傳慢指針＋1，因為要的是長度
（main function）
- 設定一個base case如果傳進來的list是null或length 0回傳0
- 設定一個指針變數first等於0
- 開始for loop迴圈（i=0; i小於list的length; i++）
3.1 如果first指針的位置的值不等於i指針位置的值時，first先走一格之後寫入，i位置的值（first是慢指針，i是快指針）
4.結束for loop，代表遍歷list結束，回傳慢指針first+1的數值，因為答案要的是長度
