搞定了...LZ给分吧!!! public int Test(int[] a){ int i=(1+a.length+1)*(a.length+1)/2; int k=0; for(int n=1;n<=a.length;n++)k+=a[n]; return i-k; }
稍微修正一下.. 第4行 public int Test(int[] a){ int i=(1+a.length+1)*(a.length+1)/2; int k=0; for(int n=0;n<a.length;n++)k+=a[n]; //这里犯了个小错-___,-||| return i-k; }
talent. good way to solve it. if it was me, I will resort String to kill this problem.
我建议排序之后查找是否连续,不过这样的话v....
起泡排序复杂度是n*n、不知道怎么弄?
LX继续
public int Test(int[] a){
int i=(1+a.length+1)*(a.length+1)/2;
int k=0;
for(int n=1;n<=a.length;n++)k+=a[n];
return i-k;
}
第4行
public int Test(int[] a){
int i=(1+a.length+1)*(a.length+1)/2;
int k=0;
for(int n=0;n<a.length;n++)k+=a[n]; //这里犯了个小错-___,-|||
return i-k;
}
good way to solve it.
if it was me, I will resort String to kill this problem.