leetcode 26:Remove Duplicates from Sorted Array (C)
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原题目:
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.
大概的意思就是,从有序数组中找到重复点,删除就好。
第一次ac的代码 (C)int removeDuplicates(int* nums, int numsSize) { if(numsSize<2) return numsSize; int num=0; int instance=*nums; for(int n=1;n<numsSize-num;) { if(*(nums+n)==instance) { num++; for(int m=n;m<numsSize-num;m++) { *(nums+m)=*(nums+m+1); } } else { instance=*(nums+n); n++; } } return numsSize-num;}
其实思路非常简单,就是一次循环,找到重复的点就进行平移。
可是效率非常的差
等我优化一下看看。。。
看下优化后的代码
int removeDuplicates(int* nums, int numsSize) { if(numsSize<2) return numsSize; int num=-1; int instance=*nums; for(int n=0;n<numsSize;n++) { if(*(nums+n)==instance) { num++; } else { instance=*(nums+n); *(nums+n-num)=instance; } } return numsSize-num;}
效率提升还是非常的高的
思路改变也是非常简单,就是把不重复的元素往前扔就好了,不需要整体移动。
但是明显还有优化的空间
待我看看其他大神的代码。
大概看了一下,基本上写的一样,但是比较简洁,少了1ms。
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