LeetCode Java First 400 题解-026

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本文介绍了一个简单的算法，用于从已排序的数组中去除重复元素，确保每个元素只出现一次，并返回处理后数组的新长度。算法在原地操作，不使用额外空间，符合常数内存约束。通过遍历数组，比较相邻元素，将非重复元素移至数组前端，实现高效去重。

Remove Duplicates from Sorted Array Easy

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.

public int removeDuplicates(int[] nums) {

    if (nums == null || nums.length == 0)

        return 0;

    int temp = nums[0], cur;

    int count = 1;

    for (int i = 1; i < nums.length; ++i) {

        cur = nums[i];

        if (temp != cur) {

            nums[count++] = cur;

        }

        temp = cur;

    }

    return count;

}

思路：顺序找即可，先检查为空情况。如果不空，保留第一元素（计数1），顺序处理后续元素，如遇与当前不同则记入原数组头部并计数。