[Leetcode] 681. Next Closest Time 解题报告
Posted 魔豆Magicbean
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题目:
Given a time represented in the format "HH:MM", form the next closest time by reusing the current digits. There is no limit on how many times a digit can be reused.
You may assume the given input string is always valid. For example, "01:34", "12:09" are all valid. "1:34", "12:9" are all invalid.
Example 1:
Input: "19:34" Output: "19:39" Explanation: The next closest time choosing from digits 1, 9, 3, 4, is 19:39, which occurs 5 minutes later. It is not 19:33, because this occurs 23 hours and 59 minutes later.
Example 2:
Input: "23:59" Output: "22:22" Explanation: The next closest time choosing from digits 2, 3, 5, 9, is 22:22. It may be assumed that the returned time is next day's time since it is smaller than the input time numerically.
思路:
首先计算出time中所包含的digits所形成的set,然后再计算出四个指针,分别指向hour的第一位和第二位,minute的第一位和第二位。接着看增加munite后是否会溢出(溢出有两种可能性,一种是超出分钟的最大范围59,另一种是指针超出了set的范围)。如果munite溢出了,则我们置munite置为最小,并且再增加hour。在增加hour的时候,如果发现hour也溢出了(hour溢出的可能性也有两种,一种是超出小时的最大范围23,另一种是指针超出了set的范围),则我们将hour也置为最小。
代码:
class Solution
public:
string nextClosestTime(string time)
set<char> st;
st.insert(time[0]), st.insert(time[1]), st.insert(time[3]), st.insert(time[4]);
vector<int> vec;
for (auto it = st.begin(); it != st.end(); ++it)
vec.push_back(*it - '0');
int h_first = distance(st.begin(), st.find(time[0]));
int h_second = distance(st.begin(), st.find(time[1]));
int m_first = distance(st.begin(), st.find(time[3]));
int m_second = distance(st.begin(), st.find(time[4]));
bool increase = false;
getNextMinute(vec, m_first, m_second, increase);
if (increase)
getNextHour(vec, h_first, h_second);
string ret;
ret += vec[h_first] + '0';
ret += vec[h_second] + '0';
ret += ":";
ret += vec[m_first] + '0';
ret += vec[m_second] + '0';
return ret;
private:
void getNextMinute(vector<int> &vec, int &m_first, int &m_second, bool &increase)
if (++m_second == vec.size())
m_second = 0;
if (++m_first == vec.size() || vec[m_first] >= 6) // overflow
increase = true;
m_first = 0;
void getNextHour(vector<int> &vec, int &h_first, int &h_second)
if (++h_second == vec.size())
h_second = 0;
if (++h_first == vec.size() || 10 * vec[h_first] + vec[h_second] >= 24)
h_first = 0;
else
if (10 * vec[h_first] + vec[h_second] >= 24) // overflow
h_first = 0;
h_second = 0;
;
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