刷題日記（13）Remove Duplicates from Sorted Array

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] <= 104
• nums is sorted in ascending order.

解題思路：

1. “in-place” 所以用雙指針，而且條件說list已經排好了
2. 一根指針讀，一根指針寫
3. 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
4. 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
5. 最後回傳慢指針＋1，因為要的是長度

（main function）

1. 設定一個base case如果傳進來的list是null或length 0回傳0
2. 設定一個指針變數first等於0
3. 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