标签：ret space i++ using duplicate memory pass ted enc
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 ofnumsbeing modified to 0,1,2,3,4 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:
// 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]);
}
题目大意：
给定一个已排序数组，删除数组中重复的元素。返回操作后数组的大小。
理 解 ：
设置 i，j 作为“快”“慢”指针，从下标0的元素与后面的元素比较，找到最近不等的元素 j 复制到下标1位置，i 指向下标1 ，从j往后找到最近不等 i 指向的元素，复制下标2位置，依次类推。
注意 边界条件的处理。预先判断数组大小为0，1 的情况。其次，防止 j 访问数组外的元素。
代 码 C++：
class Solution { public: int removeDuplicates(vector<int>& nums) { int n = nums.size(); if(n==0) return 0; if(n==1) return 1; int count=1,i,j; for(i=0,j=1;i<n && j<n;++i){ while(j<n && nums[i]==nums[j]){ ++j; } if(j>=n) break; nums[i+1] = nums[j]; ++j; ++count; } return count; } };
运行结果：
letecode [26] - Remove Duplicates from Sorted Array
标签：ret space i++ using duplicate memory pass ted enc
原文地址：https://www.cnblogs.com/lpomeloz/p/10950386.html
