365天挑战LeetCode1000题——Day 092 能否连接形成数组 可被三整除的最大和 令牌放置

1640. 能否连接形成数组

代码实现（哈希表）

class Solution {
public:
    bool canFormArray(vector<int>& arr, vector<vector<int>>& pieces) {
        unordered_map<int, int> dict;
        int n = pieces.size();
        for (int i = 0; i < n; i++) {
            dict[pieces[i][0]] = i;
        }
        vector<int> curArr;
        int curPtr = 0;
        for (auto num : arr) {
            // cout << num << " " << curPtr << endl;
            if (curArr.empty() || curPtr == curArr.size()) {
                if (!dict.count(num)) {
                    return false;
                }
                curArr = pieces[dict[num]];
                curPtr = 1;
            }
            else if (curArr[curPtr] != num) {
                return false;
            }
            else {
                curPtr++;
            }
        }
        return true;
    }
};

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1262. 可被三整除的最大和

代码实现（动态规划）

class Solution {
public:
    int maxSumDivThree(vector<int>& nums) {
         int N = nums.size();
        int remainder[3] = {0};
        int a,b,c;
        for(int i = 0; i < N; i++){
            a = remainder[0] + nums[i];
            b = remainder[1] + nums[i];
            c = remainder[2] + nums[i];
            remainder[a%3] = max(remainder[a%3], a);
            remainder[b%3] = max(remainder[b%3], b);
            remainder[c%3] = max(remainder[c%3], c);
        }
        return remainder[0];
    }
};

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948. 令牌放置

代码实现（贪心）

class Solution {
public:
    int bagOfTokensScore(vector<int>& tokens, int power) {
        sort(tokens.begin(), tokens.end());
        if (tokens.empty() || power < tokens[0]) return 0;
        int l = 0, r = tokens.size() - 1;
        int score = 0;
        while (l <= r) {
            if (power >= tokens[l]) {
                power -= tokens[l];
                score++;
                l++;
            }
            else if (l == r) return score;
            else {
                power += tokens[r--] - tokens[l++];
            }
        }
        return score;
    }
};

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