[LeetCode] 948. Bag of Tokens 令牌包

标签：value   令牌   com   xpl   pre   log   pow   ati   html   

You have an initial power?P, an initial score of?0?points, and a bag of tokens.

Each token can be used at most once, has a value?token[i], and has potentially two ways to use it.

• If we have at least?token[i]?power, we may play the token face up, losing?token[i]?power, and gaining?1?point.
• If we have at least?1?point, we may play the token face down, gaining?token[i]?power, and losing?1?point.

Return the largest number of points we can have after playing any number of tokens.

Example 1:

Input: tokens = [100], P = 50
Output: 0

Example 2:

Input: tokens = [100,200], P = 150
Output: 1

Example 3:

Input: tokens = [100,200,300,400], P = 200
Output: 2

Note:

1. tokens.length <= 1000
2. 0 <= tokens[i] < 10000
3. 0 <= P < 10000

解法一：

class Solution {
public:
    int bagOfTokensScore(vector<int>& tokens, int P) {
        int res = 0, cur = 0, n = tokens.size(), i = 0, j = n - 1;
        sort(tokens.begin(), tokens.end());
        while (i <= j) {
            while (i <= j && tokens[i] <= P) {
                P -= tokens[i++];
                res = max(res, ++cur);
            }
            if (i > j || cur == 0) break;
            --cur;
            P += tokens[j--];
        }
        return res;
    }
};

我们也可以换一种写法，不用 while 套 while，而是换成赏心悦目的 if ... else 语句，其实也没差啦，参见代码如下：

解法二：

class Solution {
public:
    int bagOfTokensScore(vector<int>& tokens, int P) {
        int res = 0, cur = 0, n = tokens.size(), i = 0, j = n - 1;
        sort(tokens.begin(), tokens.end());
        while (i <= j) {
            if (P >= tokens[i]) {
                P -= tokens[i++];
                res = max(res, ++cur);
            } else if (cur > 0) {
                --cur;
                P += tokens[j--];
            } else {
                break;
            }
        }
        return res;
    }
};

我们也可以使用递归来做，使用一个子函数 helper，将i和j当作参数输入，其实原理跟上的方法一摸一样，不难理解，参见代码如下：

解法三：

class Solution {
public:
    int bagOfTokensScore(vector<int>& tokens, int P) {
        sort(tokens.begin(), tokens.end());
        return helper(tokens, P, 0, (int)tokens.size() - 1, 0);
    }
    int helper(vector<int>& tokens, int P, int i, int j, int cur) {
        if (i > j) return cur;
        int res = cur;
        if (tokens[i] <= P) {
            res = max(res, helper(tokens, P - tokens[i], i + 1, j, cur + 1));
        } else if (cur > 0) {
            res = max(res, helper(tokens, P + tokens[j], i, j - 1, cur - 1));
        }
        return res;
    }
};

Github 同步地址:

https://github.com/grandyang/leetcode/issues/948

参考资料：

https://leetcode.com/problems/bag-of-tokens/

https://leetcode.com/problems/bag-of-tokens/discuss/383249/Java-Solution-With-Explanation

https://leetcode.com/problems/bag-of-tokens/discuss/197696/C%2B%2BJavaPython-Greedy-%2B-Two-Pointers

LeetCode All in One 题目讲解汇总(持续更新中...)

[LeetCode] 948. Bag of Tokens 令牌包

标签：value   令牌   com   xpl   pre   log   pow   ati   html   

原文地址：https://www.cnblogs.com/grandyang/p/13092555.html