題目鏈接
tag:

• Easy;
• Two Pointer;

question
??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]);
}

思路：
??這道題要我們從有序數(shù)組中去除重復項。那么這道題的解題思路是，我們使用快慢指針來記錄遍歷的坐標，最開始時兩個指針都指向第一個數(shù)字，如果兩個指針指的數(shù)字相同，則快指針向前走一步，如果不同，則兩個指針都向前走一步，這樣當快指針走完整個數(shù)組后，慢指針當前的坐標加1就是數(shù)組中不同數(shù)字的個數(shù)，代碼如下：

class Solution {
public:
    int removeDuplicates(vector<int>& nums) {
        if (nums.empty()) return 0;
        int slow = 0, fast = 0, n = nums.size();
        while (fast < n) {
            if (nums[slow] == nums[fast]) ++fast;
            else nums[++slow] = nums[fast++];
        }
        return slow+1;
    }
};