文章目錄

• 1. 題目描述：
• 2. 思路分析：
• 3. Java程式碼：

1. 題目描述：

Given a sorted array nums, remove the duplicates in-place such that each element appear only once and return the new length.

Do not allocate extra space for another array, you must do this by modifying the input array in-place

Example 1:

Given nums = [1,1,2],
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:

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

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 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]);
}

2. 思路分析：

題目的意思是給定一個有序陣列，其中含有重複數字，最後返回非重複數字個數(length)，並且要求不要使用額外的儲存空間，且最後陣列的前length個數是去掉重複的。

我們可以使用快慢指標來記錄遍歷的座標，最開始時兩個指標都指向第一個數字，如果兩個指標指的數字相同，則快指標向前走一步，如果不同，則兩個指標都向前走一步，這樣當快指標走完整個陣列後，慢指標當前的座標加1就是陣列中不同數字的個數。注意還得交換陣列中的值，見下列程式碼。

3. Java程式碼：

原始碼：見我GiHub主頁

程式碼：

public static int removeDuplicates(int[] nums) {
    if (nums.length == 0) {
        return 0;
    }
    int preIndex = 0;
    for (int i = 1; i < nums.length; i++) {
        if (nums[i] != nums[preIndex]) {
            // 移動元素值
            nums[++preIndex] = nums[i];
        }
    }
    return preIndex + 1;
}