LeetCode算法入門- Remove Duplicates from Sorted Array -day21

題目描述
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:

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:

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.

題目分析
將一個數組重復的元素去掉，并返回數組長度。

Java實現
定義兩個指針，i指針從0開始，j從1開始，如果nums[i]與nums[j]相等，則j向后移動，移到nums[i]與nums[j]不相等，這是才將
nums[++i] = nums[j]，最后返回i+1.

class Solution {public int removeDuplicates(int[] nums) {if(nums.length == 0) return 0;int i = 0;for(int j = 1; j < nums.length; j++){if(nums[j] != nums[i]){i++;nums[i] = nums[j];}}return i+1;} }

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