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?
href: https://oj.leetcode.com/problems/single-number/
解题思路:两个相同的数异或为0,然后分为正负两种情况考虑,就得到线性的解法。
class Solution {
public:
int singleNumber(int A[], int n) {
int zres=0;
int fres=0;
for(int i=0;i<n;i++){
if(A[i]>=0){
zres ^=A[i];
}else{
fres ^=-A[i];
}
}
if(0==zres&&0==fres){
return 0;
}else if(zres!=0){
return zres;
}else{
return -fres;
}
}
};
public class Solution {
public int singleNumber(int[] A) {
int zres=0;
int fres=0;
for(int i=0;i<A.length;i++){
if(A[i]>=0){
zres ^=A[i];
}else{
fres ^=-A[i];
}
}
if(0==zres&&0==fres){
return 0;
}else if(zres!=0){
return zres;
}else{
return -fres;
}
}
}