leetCode-数组：Remove Duplicates from Sorted Array

Remove Duplicates from Sorted Array：从排列后的数组中删除重复元素

考察数组的基本操作：

class Solution {public int removeDuplicates(int[] nums) {if (nums==null || nums.length==0)return 0;int index = 1;for(int i =1; i<nums.length; i++){if(nums[i]!=nums[i-1]){nums[index] = nums[i];index++;}}return index;}public static void main(String[] args) {int[] arr = { 1, 2, 2, 3, 3 };arr = removeDuplicates(arr);System.out.println(arr.length);
}
}

Remove Duplicates from Sorted Array II (Java)

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].

最多允许两个重复，输出结果数组。

解法1：当counter是2时，就直接跳过即可，否则说明元素出现次数没有超，继续放入结果数组，若遇到新元素则重置counter。总体算法只需要扫描一次数组，所以时间上是O(n)，空间上只需要维护一个index和counter，所以是O(1)。

public int removeDuplicates(int[] A) {if(A==null || A.length==0)return 0;int idx = 0;int count = 0;for(int i=1;i<A.length;i++){if(A[i]==A[i-1]){count++;if(count>=3)continue;}else{count = 1;}A[idx++]=A[i];}return idx;
}

解法二：

public class Solution {public int removeDuplicates(int[] A) {if (A.length <= 2)return A.length;int prev = 1; // point to previousint curr = 2; // point to currentwhile (curr < A.length) {if (A[curr] == A[prev] && A[curr] == A[prev - 1]) {curr++;} else {prev++;A[prev] = A[curr];curr++;}}return prev + 1;}
}

转载于:https://www.cnblogs.com/DLlearning/p/7879711.html

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Remove Duplicates from Sorted Array II (Java)

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