Leetcode - 26 - Remove Duplicates from Sorted Array

最新推荐文章于 2021-02-09 11:30:46 发布

原创最新推荐文章于 2021-02-09 11:30:46 发布 · 109 阅读

0 ·

CC 4.0 BY-SA版权

algorithms 专栏收录该内容

29 篇文章

订阅专栏

本文介绍如何在不使用额外空间的情况下从有序数组中去除重复元素，并保持原数组只包含唯一元素的方法。通过几个简洁的代码示例，展示了如何在不同编程语言如Java和C++中实现这一功能。

Given a sorted array nums, remove the duplicates in-place such 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]);
}

题目理解：输入字符串（引用），可另外使用O（1）空间更改此字符串使没有重复字符。

由于在同一字符串上操作，修改后的字符串可理解为前k（=newStr.len）位为有效位的字符串，修改前后调用字符串的引用不变，因此返回k便可知修改后的字符串。

Solutions：

Java:

save the index of the number to be compared

compare in the loop and change the number in location 'l' when it differs from nums[save]

class Solution {
    public int removeDuplicates(int[] nums) {
        int l = 1; // locate the index of the place to be changed and count the effective digits
        if (nums == null)
            return -1;  
        int save = 0; //index of the current saved number
        
        for (int i = 1; i < nums.length; i++) {
            if (nums[i] != nums[save]) {
                nums[l++] = nums[i];
                save = i;  
            }
        }
        return l;
    }
}

set the location to be saved in to i

public int removeDuplicates(int[] nums) {
    int i = 0;
    for (int n : nums)
        if (i == 0 || n > nums[i-1])
            nums[i++] = n;
    return i;
}

public int removeDuplicates(int[] nums) {
    int i = nums.length > 0 ? 1 : 0;
    for (int n : nums)
        if (n > nums[i-1])
            nums[i++] = n;
    return i;
}

// https://leetcode.com/problems/remove-duplicates-from-sorted-array/discuss/11780/5-lines-C%2B%2BJava-nicer-loops

class Solution {
    public:
    int removeDuplicates(int A[], int n) {
        if(n < 2) return n;
        int id = 1;
        for(int i = 1; i < n; ++i) 
            if(A[i] != A[i-1]) A[id++] = A[i];
        return id;
    }
};
// https://leetcode.com/problems/remove-duplicates-from-sorted-array/discuss/11757/My-Solution-%3A-Time-O(n)-Space-O(1)

C++:

count the number that has appeared (saved in count)

the correct place for the first appear number is i-count

int count = 0;
for(int i = 1; i < n; i++){
    if(A[i] == A[i-1]) count++;
    else A[i-count] = A[i];
}
return n-count;
// https://leetcode.com/problems/remove-duplicates-from-sorted-array/discuss/11782/Share-my-clean-C%2B%2B-code

set the location to be saved in to i

int removeDuplicates(vector<int>& nums) {
    int i = 0;
    for (int n : nums)
        if (!i || n > nums[i-1])
            nums[i++] = n;
    return i;
}

int removeDuplicates(vector<int>& nums) {
    int i = !nums.empty();
    for (int n : nums)
        if (n > nums[i-1])
            nums[i++] = n;
    return i;
}

// https://leetcode.com/problems/remove-duplicates-from-sorted-array/discuss/11780/5-lines-C%2B%2BJava-nicer-loops