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题目

26. Remove Duplicates from Sorted Array
Given a sorted array nums, remove the duplicates in-place such that each element appears only once and returns 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.

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 a 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]);
}

Example 1:

Input: nums = [1,1,2]
Output: 2, nums = [1,2]
Explanation: 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:

Input: nums = [0,0,1,1,1,2,2,3,3,4]
Output: 5, nums = [0,1,2,3,4]
Explanation: 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.

Constraints:

0 <= nums.length <= 3 * 104
-104 <= nums[i] <= 104
nums is sorted in ascending order.

用两个指针解法

判断边界，如果数组为空则返回0；
记录上一个数字pre，和目前没有重复的位置index；
遍历数组一直到找出跟上个数字不等，则把当前数字赋值给下个index，并用当前数字替换pre；

class Solution {public int removeDuplicates(int[] nums) {// check edgeif (nums == null || nums.length == 0) {return 0;}int pre = nums[0];int index = 1;for (int k = 1; k < nums.length; k++) {if (nums[k] != pre) {nums[index] = nums[k];index++;pre = nums[k];}}return index;}
}

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算法：从有序数组中移除重复的数据26. Remove Duplicates from Sorted Array

题目

用两个指针解法

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