Description
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.
Solution
Two-pointers
由于是sorted array斤程，遍歷時只需要將當前元素新array的tail元素進行比較即可伟骨。
public class Solution {
public int removeDuplicates(int[] nums) {
if (nums == null || nums.length == 0) {
return 0;
}
int i = 0;
// since the index of n is not needed, we could use the beautiful iterator
for (int n : nums) {
if (i == 0 || n > nums[i - 1]) { // because it's sorted
nums[i++] = n; // no need to swap value, just override it
}
}
return i;
}
}
