[leetcode] Remove Duplicates from Sorted Array

最新推荐文章于 2022-04-06 16:22:53 发布

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本文介绍了一种方法，在不使用额外空间的情况下，去除有序数组中的重复元素，并返回新的数组长度。通过迭代和条件判断，实现原地操作，确保空间复杂度为常数。

From : https://leetcode.com/problems/remove-duplicates-from-sorted-array/

Given a sorted array, 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 in place with constant memory.

For example,
Given input array 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 new length.

class Solution {
public:
    int removeDuplicates(vector<int>& nums) {
        if(nums.size()==0) return 0;
        for(vector<int>::iterator i=nums.begin(); i!=nums.end()-1; i++) {
            while(i!=nums.end() && i+1!=nums.end() && *i == *(i+1)) {
                nums.erase(i+1);
            }
            if(i == nums.end()) break;
        }
        return nums.size();
    }
};

public class Solution {
    public int removeDuplicates(int[] nums) {
        if(null == nums) {
            return 0;
        }
        int l = nums.length;
        if(l < 2) {
            return l;
        }
        int kii = nums[0];
        int ki = nums[1];
        int indice = 2;
        for(int i=2; i<l; ++i) {
            int n = nums[i];
            if(n != kii) {
                nums[indice++] = n;
            }
            kii = ki;
            ki = n;
        }
        return indice;
    }
}