leetcode數組中查找數，[leetcode] Single Number 查找數組中的單數

2023-10-17 阅读 28 评论 0

摘要：Given an array of integers, every element appears?twice?except for one. Find that single one. Note: Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory? 題目大意：一個整形數組，除了一個數例

Given an array of integers, every element appears?twice?except for one. Find that single one.

Note:
Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory?

題目大意：一個整形數組，除了一個數例外，其他每個數都會重復兩遍，找出這個單數

leetcode數組中查找數。

int compare(const void *a, const void *b)
{return (*(int*)a - *(int*)b); 
}int singleNumber(int* nums, int numsSize) {int i = 0;qsort(nums, numsSize, sizeof(int), &compare);if(numsSize == 1)return nums[0];if(nums[0] != nums[1])return nums[0];for(i = 3; i < numsSize-1; i=i+2){if(nums[i] != nums[i-1])return nums[i-1];}return nums[numsSize-1];
}

運行時間12ms，其他人最快8ms

排序的時候其實就可以找出這個單數了，而上面的代碼使用了庫函數先排序，再循環查找，浪費了時間

原文链接：https://hbdhgg.com/4/143584.html

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