【LeetCode】解题26：Remove Duplicates from Sorted Array

Problem 26: Remove Duplicates from Sorted Array [Easy]

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]);
}

来源：LeetCode

解题思路

• 使用一个指针index指向需要修改的位置，nums[0]~nums[index-1]代表已经检测过的未重复的数字。
• 遍历数组，跳过重复的数字，将未重复的数字填入nums[index]，指针index向后移一位。
• 数组个数小于2时直接返回。

时间复杂度O(n)。

要点：双指针

Solution (Java)

class Solution {
    public int removeDuplicates(int[] nums) {
        int N = nums.length;
        if (N < 2) return N;
        int index = 1;
        int last = nums[0];
        for(int i = 1; i < N; i++){
            while(i < N && nums[i] == last) i++;
            if(i == N) break;
            last = nums[i];
            nums[index] = nums[i];
            index++;
        }
        return index;
    }
}

修改过程

• 跳过重复数字时忘记加上i < N的判断，导致数组越界。