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- Given a sorted array nums, 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 by modifying the input array in-place with O(1) extra memory.
- Example 1:
- Given nums = [1,1,2],
- Your function should return length = 2, with the first two elements of nums being 1 and 2 respectively.
- It doesn't matter what you leave beyond the returned length.
- Example 2:
- Given nums = [0,0,1,1,1,2,2,3,3,4],
- Your function should return length = 5, with the first five elements of nums being modified to 0, 1, 2, 3, and 4 respectively.
- It doesn't matter what values are set beyond the returned length.
- 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 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]);
- }
复制代码
- class Solution {
- public int removeDuplicates(int[] nums) {
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- if(nums == null || nums.length == 0) return 0;
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- if(nums.length == 1) return 1;
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- int len = nums.length;
-
- int count = 1;
-
- for(int i = 1; i < len; i++){
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- if(nums[i-1] != nums[i]){
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- nums[count++] = nums[i];
- }
- }
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- return count;
-
- }
-
- }
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