package leetcode_1To300;

/**
 * 本代码来自 Cspiration，由 @Cspiration 提供
 * 题目来源：http://leetcode.com
 * - Cspiration 致力于在 CS 领域内帮助中国人找到工作，让更多海外国人受益
 * - 现有课程：Leetcode Java 版本视频讲解（1-900题）（上）（中）（下）三部
 * - 算法基础知识（上）（下）两部；题型技巧讲解（上）（下）两部
 * - 节省刷题时间，效率提高2-3倍，初学者轻松一天10题，入门者轻松一天20题
 * - 讲师：Edward Shi
 * - 官方网站：https://cspiration.com
 * - 版权所有，转发请注明出处
 */
public class _26_RemoveDuplicatesfromSortedArray {
    /**
     * 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.

     case : [1,1,2,2,3,4,5,6]
             1,2,3,4,5,6
                 c
                     i
     result : [1,2,3,4,5,6]

     time : O(n);
     space : O(1);

     * @param nums
     * @return
     */
    public static int removeDuplicates(int[] nums) {
        if (nums == null || nums.length == 0) return 0;
        int count = 1;
        for (int i = 1; i < nums.length; i++) {
            if (nums[i - 1] != nums[i]) {
                nums[count++] = nums[i];
            }
        }
        return count;
    }
}