LeetCode Problem 136:Single Number
2014-12-23 19:29
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描述: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?
题目要求O(n)时间复杂度,O(1)空间复杂度。
思路1:初步使用暴力搜索,遍历数组,发现两个元素相等,则将这两个元素的标志位置为1,最后返回标志位为0的元素即可。时间复杂度O(n^2)没有AC,Status:Time Limit Exceed
思路2:利用异或操作。异或的性质1:交换律a ^ b = b ^ a,性质2:0 ^ a = a。于是利用交换律可以将数组假想成相同元素全部相邻,于是将所有元素依次做异或操作,相同元素异或为0,最终剩下的元素就为Single Number。时间复杂度O(n),空间复杂度O(1)
Note:
Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory?
题目要求O(n)时间复杂度,O(1)空间复杂度。
思路1:初步使用暴力搜索,遍历数组,发现两个元素相等,则将这两个元素的标志位置为1,最后返回标志位为0的元素即可。时间复杂度O(n^2)没有AC,Status:Time Limit Exceed
class Solution { public: int singleNumber(int A[], int n) { vector <int> flag(n,0); for(int i = 0; i < n; i++) { if(flag[i] == 1) continue; else { for(int j = i + 1; j < n; j++) { if(A[i] == A[j]) { flag[i] = 1; flag[j] = 1; } } } } for(int i = 0; i < n; i++) { if(flag[i] == 0) return A[i]; } } };
思路2:利用异或操作。异或的性质1:交换律a ^ b = b ^ a,性质2:0 ^ a = a。于是利用交换律可以将数组假想成相同元素全部相邻,于是将所有元素依次做异或操作,相同元素异或为0,最终剩下的元素就为Single Number。时间复杂度O(n),空间复杂度O(1)
class Solution { public: int singleNumber(int A[], int n) { //异或 int elem = 0; for(int i = 0; i < n ; i++) { elem = elem ^ A[i]; } return elem; } };
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