Leetcode-26 Remove Duplicates from Sorted Array

Remove Duplicates from Sorted Array

题目描述

Given a sorted array, remove the dupicates 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 with O(1) extra memory.

Example

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 new length.

解法1

这道题要我们从有序数组中去除重复项，并返回去重后的数组长度。那么这道题的解题思路是，我们使用快慢指针来记录遍历的坐标，最开始时两个指针都指向第一个数字，如果两个指针指的数字相同，则快指针向前走一步，如果不同，则两个指针都向前走一步，这样当快指针走完整个数组后，慢指针当前的坐标加 1 就是数组中不同数字的个数，代码如下：

Cpp

class Solution {
public:
    int removeDuplicates(vector<int>& nums) {
        if (nums.empty()) return 0;
        int n = nums.size(), pre = 0, cur = 0;
        while (cur < n) {
            if (nums[pre] == nums[cur]) ++cur;
            else nums[++pre] = nums[cur++];
        }
        return pre + 1;
    }
};

Swift

class Solution {
    func removeDuplicates(_ nums: inout [Int]) -> Int {
        if nums.isEmpty { return 0 }
        var pre = 0, cur = 0
        let n = nums.count
        while cur < n {
            if nums[pre] == nums[cur] { cur += 1 }
            else {
                pre += 1
                nums[pre] = nums[cur]
                cur += 1
            }
        }
        return pre + 1
    }
}

解法2

我们也可以用for循环来写，这里的j就是上面解法中的pre，i就是cur，所以本质上都是一样的，参见代码如下：

Cpp

class Solution {
public:
    int removeDuplicates(vector<int>& nums) {
        if (nums.empty()) return 0;
        int n = nums.size(), j = 0;
        for (int i = 0; i < n; i++) {
            if (nums[j] != nums[i]) {
                nums[++j] = nums[i];
            }
        }    
        return pre + 1;
    }
};

Swift

class Solution {
    func removeDuplicates(_ nums: inout [Int]) -> Int {
        if nums.isEmpty { return 0 }
        var j = 0
        let n = nums.count
        for i in 0..<n {
            if nums[j] != nums[i] {
                j += 1
                nums[j] = nums[i]
            }
        }
        return pre + 1
    }
}