PAT 1101. Quick Sort (25)

题意：给你一个长度为n的数组，问你有几个数满足左边的都比它小，右边的都比它大，并输出他们。
思路：自从11月退役后好久好久没做题了。。不会做题了，，这么水的题，只要找到每个数左边的最大值和右边的最小值跟自己比较一下就行了，我上来居然想到用线段树、、、、其实从头和从尾巴扫一遍记录下最值就可以了。。。

代码：#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn = 1e5+5;
const int INF = 0x3f3f3f3f;
int a[maxn], n;
int okL[maxn], okR[maxn];
int res[maxn];

int main(void)
{
while(cin >> n)
{
for(int i = 1; i <= n; i++)
scanf("%d", &a[i]);
memset(okL, 0, sizeof(okL));
memset(okR, 0, sizeof(okR));
int tmp = -INF;
for(int i = 1; i <= n; i++)
{
if(a[i] < tmp) okL[i] = 1;
tmp = max(tmp, a[i]);
}
tmp = INF;
for(int i = n; i >= 1; i--)
{
if(a[i] > tmp) okR[i] = 1;
tmp = min(tmp, a[i]);
}
int ans = 0;
for(int i = 1; i <= n; i++)
if(!okL[i] && !okR[i])
res[ans++] = a[i];
printf("%d\n", ans);
for(int i = 0; i < ans; i++)
printf("%d%c", res[i], i==ans-1 ? '\n' : ' ');
if(!ans) puts("");
}
return 0;
}好久没写线段树了，，就当复习线段树。。
代码：#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn = 1e5+5;
int n;
int a[maxn], A[maxn], Hash[maxn];
int treeMax[maxn*4], treeMin[maxn*4];

void build(int root, int l, int r)
{
if(l == r)
{
treeMax[root] = treeMin[root] = A[l];
return ;
}
int mid = (l+r)/2;
build(root*2, l, mid);
build(root*2+1, mid+1, r);
treeMax[root] = max(treeMax[root*2], treeMax[root*2+1]);
treeMin[root] = min(treeMin[root*2], treeMin[root*2+1]);
}

int queryMax(int root, int l, int r, int i, int j)
{
if(i <= l && j >= r)
return treeMax[root];
int mid = (l+r)/2;
int ans = -1;
if(i <= mid) ans = max(ans, queryMax(root*2, l, mid, i, j));
if(j > mid) ans = max(ans, queryMax(root*2+1, mid+1, r, i, j));
return ans;
}

int queryMin(int root, int l, int r, int i, int j)
{
if(i <= l && j >= r)
return treeMin[root];
int mid = (l+r)/2;
int ans = maxn+1;
if(i <= mid) ans = min(ans, queryMin(root*2, l, mid, i, j));
if(j > mid) ans = min(ans, queryMin(root*2+1, mid+1, r, i, j));
return ans;
}

int res[maxn];

int main(void)
{
while(cin >> n)
{
for(int i = 1; i <= n; i++)
scanf("%d", &a[i]), Hash[i] = a[i];
sort(Hash+1, Hash+1+n);
int d = unique(Hash+1, Hash+1+n)-Hash-1;
for(int i = 1; i <= n; i++)
A[i] = lower_bound(Hash+1, Hash+1+d, a[i])-Hash;
build(1, 1, n);
int tmp, ans = 0;
for(int i = 1; i <= n; i++)
{
if(i == 1)
{
tmp = queryMin(1, 1, n, 2, n);
if(tmp > A[i])
res[ans++] = a[i];
}
else if(i == n)
{
tmp = queryMax(1, 1, n, 1, n-1);
if(tmp < A[i])
res[ans++] = a[i];
}
else
{
int R = queryMin(1, 1, n, i+1, n);
int L = queryMax(1, 1, n, 1, i-1);
if(L < A[i] && R > A[i])
res[ans++] = a[i];
}
}
printf("%d\n", ans);
for(int i = 0; i < ans; i++)
printf("%d%c", res[i], i==ans-1 ? '\n' : ' ');
if(ans == 0) puts("");
}
return 0;
}