【LeetCode题目记录-10】已排序的数组去除重复元素

标签：已排序的数组去除重复元素

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 A = [1,1,2],

Your function should return length = 2, and A is now [1,2].

【分析1-原创】重点是重复的话如何用后面不重复的元素进行覆盖。

   /* position记录重复，初始为0，一旦重复出现就递增。如果不重复，且positon>0,表明前面有重复，需要交换到前面。*/
    public static int removeDuplicates(int[] A) {
        if (A == null || A.length == 0)
            return 0;
        int position = 0, counter = 1;
        for (int i = 1; i < A.length; i++) {
            if (A[i] == A[i - 1]) {
                position++;
            } else {
                if (position > 0) {
                    A[i - position] = A[i];
                }
                counter++;
            }
        }
        return counter;
    }

【分析2-非原创】只要后面的元素跟前面的不同就将它放到counter指定的位置，并把counter加1。counter初始是0。

参考：https://oj.leetcode.com/discuss/10314/my-solution-time-o-n-space-o-1

 public static int removeDuplicates(int[] A) {
        if (A == null || A.length == 0)
            return 0;
        int counter = 1;
        for (int i = 1; i < A.length; i++) {
            if (A[i] != A[i - 1])
                A[counter++] = A[i];
        }
        return counter;
    }

【LeetCode题目记录-10】已排序的数组去除重复元素

标签：已排序的数组去除重复元素

原文地址：http://blog.csdn.net/brillianteagle/article/details/39295251