[LeetCode]26. Remove Duplicates from Sorted Array 解题报告(C++)

[LeetCode]26. Remove Duplicates from Sorted Array 解题报告(C++)

题目描述

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.

题目大意

• 给定一个以及排序的数组
• 去除排序数组中的重复元素.
• 使得每个字符出现一次.返回新的长度

解题思路

方法1:

• 暴力方法:需要额外空间
• 利用 set 的有序以及唯一性质.

代码实现:

int removeDuplicates1(vector<int>& nums) {
    set<int> myset(nums.begin(), nums.end());
    nums.clear();
    auto it = myset.begin();
    while (it != myset.end()) {
        nums.push_back(*it);
        it++;
    }
    return myset.size();
}

方法2:

• 快慢指针
• 若数字相同,快指针++
• 若数字不同,则将快指针指向的值付给慢指针i+1. 并且i++,j++

代码实现:

int removeDuplicates(vector<int>& nums) {
    int size = nums.size();
    if (size <= 1) {
        return size;
    }
    int i, j;
    for (i = 0, j = 0; j < size;) {
        if (nums[i] == nums[j]) {
            j++;
        }
        else {
            nums[i+1] = nums[j];
            i++;
            j++;
        }
    }
    return i+1;
}

// 简化写法:
class Solution {
public:
    int removeDuplicates(vector<int>& nums) {
    int size = nums.size();
    if (size < 1) {
        return size;
    }
    int i, j;
    for (i = 0, j = 0; j < size;j++) {
        if(nums[i]!=nums[j]){
            nums[++i] = nums[j];
        }
    }
    return i+1;
    }/
};

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