最后一位设为INF,这样比较的时候就不用判断i和j是不是到达了k1和k2
(没有AC,最后内存超限,为什么要卡这个,,,感觉好没意思)
#include<iostream>
#include<algorithm>
using namespace std;
const int INF = 1e9 + 10;
const int maxn = 2e5 + 10;
int a[maxn], b[maxn], k1, k2;
int main()
{
scanf("%d", &k1);
for (int i = 0; i < k1; i++)scanf("%d", &a[i]);
a[k1] = INF;
scanf("%d", &k2);
for (int i = 0; i < k2; i++)scanf("%d", &b[i]);
b[k2] = INF;
int mid = (k1 + k2 - 1) / 2;//找的是答案的前一个
int cnt = 0, i = 0, j = 0;
while (cnt < mid)
{
if (a[i] < b[j])i++;
else j++;
cnt++;
}
if (a[i] < b[j])printf("%d", a[i]);//这里找最后一个
else printf("%d", b[j]);
return 0;
}