leetcode 题集

775. Global and Local Inversions

统计相邻元素组成的逆序对(local inversion)和全局逆序对的数量(global inversion)

思路:local inversion 很容易扫一遍即可,global逆序对可以用树状数组进行统计.

class Solution {
public:
    int C[];
    int lowbit(int x){
        return x&(-x);
    }
    void Insert(int id,int v){
        for(int i=id;i<;i+=lowbit(i)){
            C[i]+=v;
        }
    }
    int getsum(int id){
        int sum = ;
        for(int i=id;i>;i-=lowbit(i)){
            sum+=C[i];
        }
        return sum;
    }
    bool isIdealPermutation(vector<int>& A) {
        if(A.size()==) return true;
        int local = ,global = ;
        memset(C,,sizeof(C));
        for(int i=;i<A.size();i++){
            if(i<A.size()-&&A[i]>A[i+]) local++;
            Insert(A[i]+,);
            global+=(i+-getsum(A[i]+));
        }
        return local==global;
    }
};

674. Longest Continuous Increasing Subsequence

最长连续上升子序列

if a[i]>a[i-1]: dp[i] = dp[i-1]+1 else: dp[i] = 1

class Solution {
public:
    int findLengthOfLCIS(vector<int>& nums) {
        if(nums.size()==) return ;
        int dp[];
        dp[] = ;
        int mx = ;
        for(int i=;i<nums.size();i++){
            if(nums[i]>nums[i-]) dp[i] = dp[i-]+;
            else dp[i] = ;
            mx = max(mx,dp[i]);
        }
        return mx;
    }
};