Description

On a table are N cards, with a positive integer printed on the front and back of each card (possibly different).

We flip any number of cards, and after we choose one card.

If the number X on the back of the chosen card is not on the front of any card, then this number X is good.

What is the smallest number that is good? If no number is good, output 0

Here, fronts[i] and backs[i] represent the number on the front and back of card i.

A flip swaps the front and back numbers, so the value on the front is now on the back and vice versa.

Example:

Input: fronts = [1,2,4,4,7], backs = [1,3,4,1,3]
Output: 2
Explanation: If we flip the second card, the fronts are [1,3,4,4,7] and the backs are [1,2,4,1,3].
We choose the second card, which has number 2 on the back, and it isn‘t on the front of any card, so 2 is good.
 

Note:

1. 1 <= fronts.length == backs.length <= 1000.
2. 1 <= fronts[i] <= 2000.
3. 1 <= backs[i] <= 2000.

Analyse

桌子上有N張牌，正反面都印有正整數
可以翻轉任意張牌，然後選一張牌，如果這張牌背面的數字X在牌的正面沒有，這個數字X就是要輸出的答案

一張牌的正反面如果是同一個數字，那這個數字肯定不是答案，這個例子裏的1，4就不可能是答案，在剩下的元素中輸出最小的那個

fronts = [1,2,4,4,7]
backs  = [1,3,4,1,3]

寫出的第一個版本

bool isExist(int value, vector<int>& vec)
{
    vector<int>::iterator it = find(vec.begin(), vec.end(), value);
    if (it != vec.end())
    {
        return true;
    }

    return false;
}

int flipgame(vector<int>& fronts, vector<int>& backs) {
    int min = 2001;
    int tmp;

    vector<int> intersect = {};

    for (int i = 0; i < fronts.size(); i++)
    {
        if (fronts[i] == backs[i])
        {
            intersect.push_back(fronts[i]);
            continue;
        }
    }

    for (int i = 0; i < fronts.size(); i++)
    {
        if (!isExist(fronts[i], intersect))
        {
            tmp = fronts[i] <
            if (isExist(backs[i], intersect))
            {
                continue;
            }
            else
            {
                tmp = backs[i];
            }
        }
        else
        {
            if (isExist(backs[i], intersect))
            {
                tmp = fronts[i];
            }
            else
            {
                tmp = fronts[i] < backs[i] ? fronts[i] : backs[i];
            }
        }

        min = min < tmp ? min : tmp;
    }

    return min == 2001 ? 0 : min;
}
 

把代碼簡化一下

bool isExist(int value, vector<int>& vec)
{
    vector<int>::iterator it = find(vec.begin(), vec.end(), value);
    if (it != vec.end())
    {
        return true;
    }

    return false;
}

int flipgame(vector<int>& fronts, vector<int>& backs) {
    int min = 2001;
    int tmp;

    vector<int> intersect = {};

    for (int i = 0; i < fronts.size(); i++)
    {
        if (fronts[i] == backs[i])
        {
            intersect.push_back(fronts[i]);
            continue;
        }
    }

    for (int i = 0; i < fronts.size(); i++)
    {
        if (!isExist(fronts[i], intersect))
        {
            tmp = fronts[i] <
            if (isExist(backs[i], intersect))
            {
                continue;
            }
            else
            {
                tmp = backs[i];
            }
        }
        else
        {
            if (isExist(backs[i], intersect))
            {
                tmp = fronts[i];
            }
            else
            {
                tmp = fronts[i] < backs[i] ? fronts[i] : backs[i];
            }
        }

        min = min < tmp ? min : tmp;
    }

    return min == 2001 ? 0 : min;
}

繼續優化，把vector換成unordered_set，在LeetCode上就會有巨大的提升

int flipgame(vector<int>& fronts, vector<int>& backs) {
    int min = 2001;
    int tmp;

    unordered_set<int> intersect = {};

    for (int i = 0; i < fronts.size(); i++)
    {
        if (fronts[i] == backs[i])
        {
            intersect.insert(fronts[i]);
            continue;
        }
    }

    for (int i = 0; i < fronts.size(); i++)
    {
        if (intersect.count(fronts[i]) == 0)
        {
            min = min < fronts[i] ? min : fronts[i];
        }

        if (intersect.count(backs[i]) == 0)
        {
            min = min < backs[i] ? min : backs[i];
        }
    }

    return min == 2001 ? 0 : min;
}

看看LeetCode上評價最高的版本

思路是差不多的，改成了用數組存儲，有點bitmap的思想，

int flipgame(vector<int>& fronts, vector<int>& backs) {
    int res[2002] = {0};  // -1: bad. 1:exist. 
    for (int i = 0; i < fronts.size(); i++)
    {
        if (fronts[i] == backs[i])
        {
            res[fronts[i]] = -1;
        }
        else
        {
            if (res[fronts[i]] != -1)
                res[fronts[i]] = 1;
            if (res[backs[i]] != -1)
                res[backs[i]] = 1;
        }
    }

    for (int i = 0; i < 2002; i++)
    {
        if (res[i] == 1)
            return i;
    }
    return 0;
}

比如

fronts = [1,2,4,4,7]
backs  = [1,3,4,1,3]

res[1] = -1;
res[2] = 1;
res[3] = 1;
res[4] = -1;
res[7] = 1;

for循環的時候從index小的開始，遇到的第一個值為1的就是要找的值

Reference

1. set_intersection - C++ Reference

2. find - C++ Reference

[LeetCode] 822. Card Flipping Game