GuilinDev

Lc0026

05 August 2008

26 - Remove Duplicates from Sorted Array

原题概述

Given a sorted array nums, 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 by modifying the input array in-place with O(1) extra memory.

Example 1:

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Given 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 returned length.

Example 2:

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Given nums = [0,0,1,1,1,2,2,3,3,4],

Your function should return length = 5, with the first five elements of nums being modified to 0, 1, 2, 3, and 4 respectively.

It doesn't matter what values are set beyond the returned length.

Clarification:

Confused why the returned value is an integer but your answer is an array?

Note that the input array is passed in by reference, which means modification to the input array will be known to the caller as well.

Internally you can think of this:

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// nums is passed in by reference. (i.e., without making a copy)
int len = removeDuplicates(nums);

// any modification to nums in your function would be known by the caller.
// using the length returned by your function, it prints the first len elements.
for (int i = 0; i < len; i++) {
    print(nums[i]);
}

题意和分析

与第27题的解法基本相同，这道题是在一个排好序的Array中，找到有多少个非duplicate的值然后存储在index里，index则在in place的位置上。

Time：O(n)。

代码

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class Solution {
  public int removeDuplicates(int[] nums) {
    if (nums == null || nums.length == 0) {
      return 0;
    }
    int slow = 0;//最左边可以swap的位置
    for (int fast = 0; fast <= nums.length - 1; fast++) {
      if (fast < 1 || nums[fast] != nums[slow - 1]) {
        if (slow != fast) { // 优化下，自己与自己不用再赋值赋值,[1,2,3,4,5]这种情况
          nums[slow] = nums[fast];
        }
        slow++;
      }
    }
    return slow;
  }
}
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