Leet Code OJ 26. Remove Duplicates from Sorted Array [Difficulty: Easy]
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题目:
Given a sorted array, 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 in place with constant memory.
For example,
Given input array 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 new length.
翻译:
给定一个排序好的数组,就地移除重复的元素,来使得每个元素只出现一次,并且返回新的长度。
不要分配额外的空间给另一个数组,你必须完成它在原数组上,并且只使用常数级的内存。
分析:
由于需要移除重复的元素,所以必须有移动元素的操作。当遍历的i指针指向的元素与上一个元素重复时,需要采用一个指针nextEmpty来记录当前这个位置(需要被移除的位置,也就是要把后面的元素复制过来的位置),当遍历到下一个不重复的元素时,再复制到这个位置。
代码:
public class Solution { public int removeDuplicates(int[] nums) { int nextEmpty=1; for(int i=1;i<nums.length;i++){ if(nums[i]!=nums[i-1]){ nums[nextEmpty]=nums[i]; nextEmpty++; } } return nextEmpty; }}
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