26. Remove Duplicates from Sorted Array

time: O(n)

space: O(1)

class Solution {
    public int removeDuplicates(int[] nums) {
        int n = nums.length;
        int i, j = 1;
        for (i = 0; i < n ; i++) {
            while (j < n && nums[j] == nums[i]) {
                j++;
            }      
            if (j == n) {
                break;
            }
            nums[i+1] = nums[j];
        }
        return i+1;
    }
}

/*
[0,1,2,3,4,2,2,3,3,4]
                   j        
         i
         
         so the ans is i+1
*/

more easier way

because when number are different, set number to front, so...

這的 i, 就像上一個做法的 j, count 就像上一個做法的 i

class Solution {
    public int removeDuplicates(int[] nums) {
        int count = 1;
        for (int i = 1; i < nums.length; i++) {
            if (nums[i-1] != nums[i]) {
                nums[count++] = nums[i];
            }
        }
        return count;
    }
}