26. Remove Duplicates from Sorted Array

Given a sorted array *nums*, remove the duplicates [**in-place**](https://en.wikipedia.org/wiki/In-place_algorithm) 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:**

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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:**
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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:
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// 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:本题要求在原来数组的基础上,不许开辟新的内存空间,删除重复的数字。可以建立指针,在遍历过程中,分别指向非重复的数组的末位,指向当前遍历的位置,以及该位置的数字。代码如下:
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class Solution {
public:
int removeDuplicates(vector<int>& nums) {
if (nums.size() == 1) return 1;
if (nums.size() == 0) return 0;

int len = 1, cur = 0, cur_nums_val = nums[0];
while (cur < nums.size() - 1) {

if (cur_nums_val != nums[cur + 1]) {
cur_nums_val = nums[cur + 1];
if (nums[len - 1] != nums[cur + 1]) {
nums[len] = nums[cur + 1];
len++;
}
}

cur++;
}

return len;
}
};
Res. Runtimes 16 ms, Ranking 99.95%. 其实,变量cur_nums_val是不必要的,简洁的写法如下:
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class Solution {
public:
int removeDuplicates(vector<int>& nums) {

if (nums.size() == 1) return 1;
if (nums.size() == 0) return 0;

int len = 1, cur = 0;

while (cur < nums.size() - 1) {
if (nums[cur] != nums[cur + 1] && nums[len - 1] != nums[cur + 1]) {
nums[len] = nums[cur + 1];
len++;
}
cur++;
}

return len;
}
};