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 <= fronts.length == backs.length <= 1000.1 <= fronts[i] <= 2000.1 <= backs[i] <= 2000.
这道题刚开始的时候博主一直没看懂题意，不知所云，后来逛了论坛才总算弄懂了题意，说是给了一些正反都有正数的卡片，可以翻面，让我们找到一个最小的数字，在卡的背面，且要求其他卡正面上均没有这个数字。简而言之，就是要在backs数组找一个最小数字，使其不在fronts数组中。我们想，既然不能在fronts数组中，说明卡片背面的数字肯定跟其正面的数字不相同，否则翻来翻去都是相同的数字，肯定会在fronts数组中。那么我们可以先把正反数字相同的卡片都找出来，将数字放入一个HashSet，也方便我们后面的快速查找。现在其实我们只需要在其他的数字中找到一个最小值即可，因为正反数字不同，就算fronts中其他卡片的正面还有这个最小值，我们可以将那张卡片翻面，使得相同的数字到backs数组，总能使得fronts数组不包含有这个最小值，就像题目中给的例子一样，数字2在第二张卡的背面，就算其他卡面也有数字2，只要其不是正反都是2，我们都可以将2翻到背面去，参见代码如下：
class Solution {
public:
int flipgame(vector<int>& fronts, vector<int>& backs) {
int res = INT_MAX, n = fronts.size();
unordered_set<int> same;
for (int i = 0; i < n; ++i) {
if (fronts[i] == backs[i]) same.insert(fronts[i]);
}
for (int front : fronts) {
if (!same.count(front)) res = min(res, front);
}
for (int back : backs) {
if (!same.count(back)) res = min(res, back);
}
return (res == INT_MAX) ? 0 : res;
}
};
参考资料：
https://leetcode.com/problems/card-flipping-game/
LeetCode All in One 题目讲解汇总(持续更新中…)
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