Remove Duplicates from Sorted Array (E)
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]);
}
题意
给定一个有序数组,要求将数组中不重复的元素依次排在数组前部,并返回不重复元素的个数。
思路
因为数组有序,直接遍历处理即可。
代码实现
class Solution {
public int removeDuplicates(int[] nums) {
int p = 0;
for (int i = 0; i < nums.length; i++) {
if (i > 0 && nums[i] == nums[i - 1]) {
continue;
}
nums[p++] = nums[i];
}
return p;
}
}
本文介绍了一种在原地修改输入数组,去除重复元素并返回新长度的方法。通过遍历数组,只保留不重复的元素,实现了O(1)额外内存使用的要求。适用于已排序的数组,展示了具体的代码实现。
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