最长递增子序列问题【C++】

**问题:**最长递增子序列问题主要分为了两类,即最长连续递增子序列的求解,以及最长递增子序列的求解(不一定要连续)。求解过程总结如下:
算法标签:动态规划、深度优先搜索、二分查找
代码:01_dp求解最长连续递增子序列长度

#include
#include
using namespace std;
const int maxN = 1e5+9;
int dp[maxN];   // dp[i]表示以第i个元素作为最后元素的最长连续递增子序列的的长度
int res = -1;
int main(){
    int n;
    int arr[maxN];
    cin >> n;
    for( int i=1; i<=n; i++ ){
        cin >> arr[i];
    }
    dp[1] = 1;
    for( int i=2; i<=n; i++ ){
        if(arr[i] > arr[i-1]){
            dp[i] = dp[i-1] + 1;
        }else{
            dp[i] = 1;
        }
        res = max(res, dp[i]);
    }
    cout << res;
    return 0;
}

代码:02_dp求最长递增子序列长度(n*n)

#include
#include
using namespace std;
const int maxN = 1e5+9;
int dp[maxN];   // dp[i]表示以第i个元素作为最后元素的最长递增子序列的长度
int res = 1;
int main(){
    int n;
    int arr[maxN];
    cin >> n;
    for( int i=1; i<=n; i++ ){
        cin >> arr[i];
    }
    for( int i=1; i<=n; i++ ){
        dp[i] = 1;
        for( int j=1; j<=i-1; j++ ){
            if(arr[i] > arr[j]){
                dp[i] = max(dp[i], dp[j]+1);
            }
        }
        res = max(res, dp[i]);
    }
    cout << res;
    return 0;
}

代码:03_dp求最长递增子序列长度(n*logn)

#include
using namespace std;

const int maxN = 1e5 + 9;
int n, res;
int dp[maxN];   // 存储当前伪最长上升子序列[与真实最长上升子序列的长度相同]
int main(){
    int arr[maxN];
    cin >> n;
    for( int i=1; i<=n; i++ ){
        cin >> arr[i];
    }
    dp[1] = arr[1];
    res = 1;
    for( int i=2; i<=n; i++ ){
        if(arr[i] > dp[res]){
            dp[++res] = arr[i];
        }else{
            int tmp = lower_bound(dp+1, dp+res+1, arr[i]) - dp;
            dp[tmp] = arr[i];
        }
    }
    cout << res;
    return 0;
}

代码:04_dfs求解最长递增子序列

#include
#include
#include
using namespace std;
int dp[100009];
int res = 0;
string ans;
void dfs(int currentIndex, int previousIndex, string tmp, string str){
    if(currentIndex == str.length()){
        if(res < tmp.length()){
            ans = tmp;
            res = tmp.length();
        }
        return;
    }

    if(currentIndex == 0 || str[currentIndex] > str[previousIndex]){
        dfs(currentIndex+1, currentIndex, tmp+str[currentIndex], str);
        dfs(currentIndex+1, previousIndex, tmp, str);
    }else{
        dfs(currentIndex+1, previousIndex, tmp, str);
        dfs(currentIndex+1, previousIndex, ""+str[currentIndex], str);
    }
}
int main(){
    string str;
    cin >> str;
    dfs(0, -1, "", str);
    cout << ans;
    return 0;
}

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