LeetCode Weekly Contest 276 题解

第一题

题解：直接模拟

class Solution {
public:
    vector<string> divideString(string s, int k, char fill) {
        vector<string> res;
        int n = s.size();
        string t ="";
        for(int i=0; i<n; i++){
            t += s[i];
            if(t.size() == k){
                res.push_back(t);
                t = "";
            }
        }
        if(t.size() > 0){
            int len = k - t.size();
            for(int i=0; i<len; i++) t+= fill;
            res.push_back(t);
        }
        return res;
    }
};

第二题

题解： 模拟，逆向思维，倒着求解

class Solution {
public:
    int minMoves(int target, int maxDoubles) {
        int res = 0;
        while(target > 1){
            if(maxDoubles == 0){
                return res + target-1;
            }
            if(target%2 == 0){
                target /= 2;
                maxDoubles--;
                res++;
            }else{
                target--;
                res++;
            }
        }
        return res;
    }
};

第三题

比赛想到的是dfs，写完提交 TLE， 比赛快结束了，想到DP，比赛期间没有写出来，看其他人的提交，发现记忆化dfs可以过。

class Solution {
public:
    long long mostPoints(vector<vector<int>>& questions) {
        int n = questions.size();
        vector<long long> dp(n+1);
        for(int i=0; i<n; i++){
            dp[i+1] = max(dp[i], dp[i+1]);
            int next = min(n, i+questions[i][1]+1);
            dp[next] = max(dp[next], dp[i]+questions[i][0]); 
        }
        return dp[n];
    }
};

第四题

题解： 二分， 参考 https://leetcode-cn.com/problems/maximum-running-time-of-n-computers/solution/er-fen-da-an-by-newhar-swi2/

class Solution {
public:

bool check(int n, long long t, vector<int>& batteries){
     long long sum = 0;
     for(auto c : batteries){
         sum += min(t, (long long)c);
     }
     return (sum/t) >= n;
}
    long long maxRunTime(int n, vector<int>& batteries) {
        long long l = 1,  r = 1e15;
        while(l < r){
             long long mid = l+r+1 >>1;
             if(check(n, mid, batteries)){
                 l = mid;
             }else{
                 r = mid-1;
             }
        }
        return l;
    }
};