[LeetCode] 822. Card Flipping Game
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 <= fronts.length == backs.length <= 1000.
- 1 <= fronts[i] <= 2000.
- 1 <= backs[i] <= 2000.
Analyse
桌子上有N张牌，正反面都印有正整数
可以翻转任意张牌，然后选一张牌，如果这张牌背面的数字X在牌的正面没有，这个数字X就是要输出的答案
一张牌的正反面如果是同一个数字，那这个数字肯定不是答案，这个例子里的1，4就不可能是答案，在剩下的元素中输出最小的那个
fronts = [1,2,4,4,7]
backs = [1,3,4,1,3]
写出的第一个版本
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;
}
把代码简化一下
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;
}
继续优化，把vector换成unordered_set，在LeetCode上就会有巨大的提升
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上评价最高的版本
思路是差不多的，改成了用数组存储，有点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循环的时候从index小的开始，遇到的第一个值为1的就是要找的值
