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Problem:
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.
Analysis:
Even though this question is very simple, it could be solved very elegantly! Just as the code I have implemented below! (try to remmber it if possible). Basic idea: Use a count the record the total number distinct numbers. ------------------------------------------------------------------------- int count = 0; ------------------------------------------------------------------------- Then we scan through the array, to check if the current element is duplicate (since it‘s sorted). ------------------------------------------------------------------------- Iff it is duplicate we directly skip it. for (int i = 0; i < nums.length; i++) { if (i == 0 || nums[i] != nums[i-1]) { nums[count] = nums[i]; count++; } } Why we can do nums[i] != nums[i-1], even we are updating on the nums array. Cause (previous)count is at most equal with i-1!!! Iff there are duplicates before i-1, count must meet : count < i-1.
Solution:
public class Solution { public int removeDuplicates(int[] nums) { if (nums == null) throw new IllegalArgumentException("nums is null"); int count = 0; for (int i = 0; i < nums.length; i++) { if (i == 0 || nums[i] != nums[i-1]) { nums[count] = nums[i]; count++; } } return count; } }
[LeetCode#26]Remove Duplicates from Sorted Array
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原文地址：http://www.cnblogs.com/airwindow/p/4784365.html
