How to delete duplicates in a sorted array in java
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Given a sorted array, you need to delete the repeating elements in place so that each element appears at most twice, returning the new length of the removed array.
Do not use extra array space, you must modify the input array in place and do it using O (1) extra space.
Example 1:
Given nums = [1, 1, 1, 1, 1, 2, 2, 3]
The function should return the new length length = 5, and the first five elements of the original array are modified to 1, 1, 2, 2, 3.
You don't need to consider the elements in the array that exceed the new length.
Example 2:
Given nums = [0jin0jing1jing1jing1jing1jinjinjingjinshengyue 3]
The function should return the new length length = 7, and the first five elements of the original array are modified to 0,0,1,1,2,3,3.
You don't need to consider the elements in the array that exceed the new length.
Answer:
1public int removeDuplicates (int [] nums) {
2 int n = nums.length
3 if (n
< 2) 4 return n; 5 int pos = 2; 6 for (int i = 2; i < n; ++i) 7 if (nums[i] != nums[pos - 2]) 8 nums[pos++] = nums[i]; 9 return pos; 10} 解析： 题目中说的是排序的数组，那这就简单了。代码比较简单就不再介绍，下面再来看一下其他的解法。 1public int removeDuplicates(int[] nums) { 2 int i = 0; 3 for (int n : nums) 4 if (i < 2 || n >Nums [I-2])
5 nums [iTunes +] = n
6 return i
7}
eight
9public int removeDuplicates (int [] nums) {
10 int n = nums.length, count = 0
11 for (int I = 2; I < n; iTunes +)
12 if (nums [I] = = nums [I-2-count])
13 count++
14 else
15 nums [I-count] = nums [I]
16 return n-count
17}
The solutions are more or less the same, and the basic ideas remain the same.
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