80 remove duplicates from sorted array 2

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

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

Your function should return length = 5, with the first five elements of nums being 1, 1, 2, 2 and 3. It doesn't matter what you leave beyond the new length.

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idx 是新数组的指针，i是nums数组的遍历指针。使用flag表示当前重复的元素个数。如果nums[i] != nums[idx],更新idx,将nums[i]存到nums[idx],同时置flag为0。如果nums[i] == nums[idx]且flag < 1，说明当前重复元素为0个，可以更新idx,将nums[i]存到 nums[idx]，同时flag加1。如果题目变为最多允许出现k次，将flag < 1改为flag < k-1即可。

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17  public int removeDuplicates(int[] nums) {
           if(nums == null || nums.length == 0) return 0;
           int n = nums.length;
           int idx = 0;
           int flag = 0;
           int tol
           for(int i = 1; i < n; i++){
                if(nums[i] != nums[idx]){
                     nums[++idx] = nums[i];
                     flag = 0;
               }else if大专栏  80 remove duplicates from sorted array 2>(nums[i] == nums[idx] && flag < 1){
                    nums[++idx] = nums[i];
                    flag++;
                }
           }
           return idx+1;
    }

2 most votes

by StefanPochmann

思路是将当前遍历元素与新数组的倒数第二个元素比较，只有n大于此元素才添加新元素。太精炼了。也可以扩展到最多允许出现k次的情况。

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7 public int removeDuplicates(int[] nums) {
    int i = 0;
    for (int n : nums)
        if (i < 2 || n > nums[i-2])
            nums[i++] = n;
    return i;
}