LWC 51：681. Next Closest Time
LWC 51：681. Next Closest Time
Problem:
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： 暴力遍历，下一个最近的时间点，只需要让时间一点点增大，并且找到第一个所有数字都被reused的情况跳出即可。
代码如下：
public String nextClosestTime(String time) {
String[] val = time.split(":");
Set<Integer> set = new HashSet<>();
int num1 = Integer.parseInt(val[0]);
set.add(num1 / 10);
set.add(num1 % 10);
int num2 = Integer.parseInt(val[1]);
set.add(num2 / 10);
set.add(num2 % 10);
int hour = num1;
int minu = num2;
minu ++;
if (minu == 60) {
hour ++;
minu = 0;
if (hour == 24) hour = 0;
}
while (!contains(hour, minu, set)) {
minu++;
if (minu == 60) {
hour ++;
minu = 0;
if (hour == 24) hour = 0;
}
}
String ans = "";
if (hour >= 0 && hour <= 9) ans = "0" + hour;
else ans += hour;
ans += ":";
if (minu >= 0 && minu <= 9) ans += "0" + minu;
else ans += minu;
return ans;
}
public boolean contains(int hour, int minu, Set<Integer> set) {
return set.contains(hour / 10) && set.contains(hour % 10) && set.contains(minu / 10) && set.contains(minu % 10);
}优化一下：
public String nextClosestTime(String time) {
String[] val = time.split(":");
Set<Integer> set = new HashSet<>();
int hour = add(set, val[0]);
int minu = add(set, val[1]);
int[] times = new int[] {hour, minu};
nxt(times);
while (!contains(times[0], times[1], set)) {
nxt(times);
}
return valid(times[0]) + ":" + valid(times[1]);
}
public void nxt(int[] time) {
int hour = time[0];
int minu = time[1];
minu ++;
if (minu == 60) {
hour ++;
minu = 0;
if (hour == 24) hour = 0;
}
time[0] = hour;
time[1] = minu;
}
public int add(Set<Integer> set, String timeStr) {
int time = Integer.parseInt(timeStr);
set.add(time / 10);
set.add(time % 10);
return time;
}
public String valid(int time) {
if (time >= 0 && time <= 9) return "0" + time;
else return time + "";
}
public boolean contains(int hour, int minu, Set<Integer> set) {
return set.contains(hour / 10) && set.contains(hour % 10) && set.contains(minu / 10) && set.contains(minu % 10);
}当然，你也可以直接搜索，因为最多只有4个数字，所以总共有4 * 4 * 4 * 4种情况，在这些情况中找出diff最小的hour和minute即可。
代码如下：
int diff = 0x3f3f3f3f;
String result = "";
int h;
int m;
public String nextClosestTime(String time) {
int[] digit = new int[4];
int tot = 0;
String[] val = time.split(":");
int hour = Integer.parseInt(val[0]);
int minu = Integer.parseInt(val[1]);
digit[tot++] = hour / 10;
digit[tot++] = hour % 10;
digit[tot++] = minu / 10;
digit[tot++] = minu % 10;
h = hour;
m = minu;
dfs(digit, 0, new int[4]);
return result;
}
void dfs(int[] digit, int i, int[] ans) {
if (i == 4) {
int hour = 10 * ans[0] + ans[1];
int minu = 10 * ans[2] + ans[3];
if (hour >= 0 && hour <= 23 && minu >= 0 && minu <= 59) {
int df = diff(hour, minu);
if (df < diff) {
diff = df;
result = valid(hour) + ":" + valid(minu);
}
}
}
else {
for (int j = 0; j < 4; ++j) {
ans[i] = digit[j];
dfs(digit, i + 1, ans);
ans[i] = -1;
}
}
}
int diff(int hour, int minu) {
int c2o = 60 * 60 - h * 60 - m;
int n2o = 60 * 60 - hour * 60 - minu;
return n2o < c2o ? c2o - n2o : c2o - n2o + 3600;
}
public String valid(int time) {
if (time >= 0 && time <= 9) return "0" + time;
else return time + "";
}剪枝优化：
int diff = 0x3f3f3f3f;
String result = "";
int h;
int m;
public String nextClosestTime(String time) {
int[] digit = new int[4];
int tot = 0;
String[] val = time.split(":");
int hour = Integer.parseInt(val[0]);
int minu = Integer.parseInt(val[1]);
digit[tot++] = hour / 10;
digit[tot++] = hour % 10;
digit[tot++] = minu / 10;
digit[tot++] = minu % 10;
h = hour;
m = minu;
dfs(digit, 0, new int[4]);
return result;
}
void dfs(int[] digit, int i, int[] ans) {
if (i == 4) {
int hour = 10 * ans[0] + ans[1];
int minu = 10 * ans[2] + ans[3];
int df = diff(hour, minu);
if (df < diff) {
diff = df;
result = valid(hour) + ":" + valid(minu);
}
}
else {
for (int j = 0; j < 4; ++j) {
ans[i] = digit[j];
if (i == 1) {
int hour = 10 * ans[0] + ans[1];
if (hour >= 0 && hour <= 23) dfs(digit, i + 1, ans);
}
else if (i == 3) {
int minu = 10 * ans[2] + ans[3];
if (minu >= 0 && minu <= 59) dfs(digit, i + 1, ans);
}
else {
dfs(digit, i + 1, ans);
}
}
}
}
int diff(int hour, int minu) {
int c2o = 60 * 60 - h * 60 - m;
int n2o = 60 * 60 - hour * 60 - minu;
return n2o < c2o ? c2o - n2o : c2o - n2o + 3600;
}
public String valid(int time) {
if (time >= 0 && time <= 9) return "0" + time;
else return time + "";
} 本文参与 腾讯云自媒体同步曝光计划，分享自作者个人站点/博客。
原始发表：2017-09-24 ，如有侵权请联系 cloudcommunity@tencent.com 删除
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