题目:
Given a sorted array nums, remove the duplicates in-place such that duplicates appeared at most twice 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 = [1,1,1,2,2,3], Your function should return length = 5, with the first five elements of nums being 1, 1, 2, 2 and 3 respectively. It doesn't matter what you leave beyond the returned length.
Example 2:
Given nums = [0,0,1,1,1,1,2,3,3], Your function should return length = 7, with the first seven elements of nums being modified to 0, 0, 1, 1, 2, 3 and 3 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]); }
public class RemoveDuplicatesSortedArrayII {
public static int removeDuplicates(int[] nums) {
//处理特殊情况
if (nums == null || nums.length == 0)
return 0;
if (nums.length == 1)
return 1;
int k = 2;
for (int i = 2; i < nums.length; i++){
if (nums[i] != nums[k - 2])
nums[k++] = nums[i];
}
return k;
}
public static void main(String[] args) {
int[] nums = {1,1,1,2,2,3};
System.out.println(removeDuplicates(nums));
}
}
博客围绕LeetCode题目展开,要求对排序数组进行原地去重,使每个元素最多出现两次,并返回新长度,不能额外分配数组空间,需在原数组上以O(1)额外内存完成操作,还给出了示例及相关说明。
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