题目:26. Remove Duplicates from Sorted Array 从有序数组中移除重复元素
难度:简单
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 ofnums
being1
and2
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 ofnums
being modified to0
,1
,2
,3
, and4
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]); }
题意解析:
给定一个排序数组,你需要在原地删除重复出现的元素,使得每个元素只出现一次,返回移除后数组的新长度。
不要使用额外的数组空间,你必须在原地修改输入数组并在使用 O(1) 额外空间的条件下完成。
解题思路一:
作为道简单题,这题几乎没什么难度,我们定义一个计数器,每当数组的第i个元素与上一个元素即i-1不同的时候我们就将i元素赋值给nums[count++]即可。
public int removeDuplicates(int[] nums) {
int len = nums.length;
if (len < 2)
return len;
int count = 1;
for (int i = 1; i < len; i++) {
if (nums[i] != nums[i-1]){
nums[count++] = nums[i];
}
}
return count;
}
提交代码之后:
Runtime: 1 ms, faster than 99.88% of Java online submissions for Remove Duplicates from Sorted Array.
Memory Usage: 40.4 MB, less than 83.52% of Java online submissions for Remove Duplicates from Sorted Array.