题目

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.

解法

思路

这道题我自己的方法是非常暴力的方法，用了三个循环，然后意料之中地超时了。看了别人的解法，第一次学到了用快慢指针来解题，时间复杂度为O(n)。刚开始先让快慢指针都指向数组中的第一个元素，判断快慢指针指向的元素是否相等，如果相等，则快指针往前走一步，慢指针不动，如果不等，慢指针先走一步，然后把快指针所指元素的值赋给慢指针所指元素，然后快指针往前走一步。

这种方法在慢指针改变数组元素的同时，可以保证快指针对数组的遍历不受影响，非常巧妙。

代码

class Solution {    public int removeDuplicates(int[] nums) {        int len = nums.length;        int j = 0; //j为慢指针        if(len == 0) return 0;        for(int i = 0; i < len; i++) { //i为快指针            if(nums[i] != nums[j])                nums[++j] = nums[i];        }        return j+1;    }}

最后附上自己的暴力解法，仅供娱乐（逃

class Solution {    public int removeDuplicates(int[] nums) {        int removeNum = 0;        int len = nums.length;        if(nums[0] == nums[len-1])            return 1;        for(int i = 0; i < len-1; i++) {            if(nums[i] > nums[i+1])                break;            while(nums[i] == nums[i+1]) {                int temp = nums[i+1];                for(int j = i+2; j < len; j++){                    nums[i+1] = nums[i+2];                    nums[len-1] = temp;                }                removeNum ++;            }        }        return len - removeNum;    }}