1029 Median
part 2, 2.3
自己解法
- 第20行的判断条件搞清楚就行
#include <iostream>
using namespace std;
#include <vector>int main()
{int n1;cin >> n1;int s1[200000];for (int i = 0; i < n1; i++)cin >> s1[i];int n2;cin >> n2;int s2[200000];for (int i = 0; i < n2; i++)cin >> s2[i];int mid = (n1 + n2 + 1) / 2;int l1 = 0, l2 = 0;for (int i = 0;; i++)if (l2 == n2 || l1 < n1 && s1[l1] < s2[l2]){if (i + 1 == mid){cout << s1[l1] << endl;break;}l1++;}else{if (i + 1 == mid){cout << s2[l2] << endl;break;}l2++;}system("pause");return 0;
}
大神解法
- 柳神
- 逻辑是一样的，但是写得更加优雅
#include <iostream>
const int maxn = 200005;
int n, m, a1[maxn], a2[maxn], cnt = 0, i, j, ans;
int main() {scanf("%d", &n);for(i = 1; i <= n; i++) scanf("%d", &a1[i]);scanf("%d", &m);for(i = 1; i <= m; i++) scanf("%d", &a2[i]);int target = (n + m + 1) / 2;i = 1, j = 1;while(i <= n && j <= m) {ans = a1[i] <= a2[j] ? a1[i++] : a2[j++];if(++cnt == target) break;}if(i <= n && cnt < target) ans = a1[i + target - cnt - 1];else if(j <= m && cnt < target) ans = a2[j + target - cnt - 1];printf("%d", ans);return 0;
}