LeetCode - Remove Duplicates from Sorted Array I && II

最新推荐文章于 2022-04-06 16:22:53 发布

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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，并允许元素最多出现两次的方法。通过双指针技巧实现，确保操作在O(n)的时间复杂度内完成。

https://leetcode.com/problems/remove-duplicates-from-sorted-array/

Given a sorted array, 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 in place with constant memory.

For example,
Given input array A = [1,1,2],

Your function should return length = 2, and A is now [1,2].

这道题也是用两个指针即可，注意，因为是sorted array，所以不需要检查current之前的，因此时间复杂度O(n)，代码如下：

    public int removeDuplicates(int[] A) {
        if(A==null || A.length==0) return 0;
        int current = 0;
        for(int i=1; i<A.length; i++){
            if(A[i]!=A[current]) A[++current] = A[i];
        }
        return (current+1);
    }

https://leetcode.com/problems/remove-duplicates-from-sorted-array-ii/

Follow up for "Remove Duplicates":
What if duplicates are allowed at most twice?

For example,
Given sorted array A = [1,1,1,2,2,3],

Your function should return length = 5, and A is now [1,1,2,2,3].

这道题跟上一道题基本一样，唯一的区别是除了要检查A[current]，还要检查A[current-1]，注意corner case，即current==0时，则无论如何都可以复制过来，只有当current>0时，才能检查A[current-1]，否则数组越界。

    public int removeDuplicates(int[] A) {
        if(A==null || A.length==0) return 0;
        int current = 0;
        for(int i=1; i<A.length; i++){
            if(A[i]!=A[current] || (current>0 && A[current-1]!=A[i]) || current==0) A[++current] = A[i];
        }
        return (current+1);
    }