[LeetCode]Minimum Time Difference(java)

我用set去重无重复则循环计算差值，而discuss则考虑用分钟hash

贴一个我的答案

public class Solution {
public int findMinDifference(List<String> timePoints) {
int result = Integer.MAX_VALUE;
Set<String> set = new HashSet<String>(timePoints);
if(set.size()<timePoints.size())
return 0;
for(int i = 0;i<timePoints.size();i++){
String[] firstTime = timePoints.get(i).split("\\:");
int firstM = (Integer.valueOf(firstTime[0]) *60+ Integer.valueOf(firstTime[1]));
for(int j = i+1;j<timePoints.size();j++){

String[] secondTime = timePoints.get(j).split("\\:");
int secondM = (Integer.valueOf(secondTime[0]) *60+ Integer.valueOf(secondTime[1]));
int val = (secondM-firstM+1440)%1440;
int min = Math.min( val,1440-val);
result = Math.min(result,min);
}
}
return result;
}
}

在贴一个hash的做法
public class Solution {
public int findMinDifference(List<String> timePoints) {
boolean[] mark = new boolean[24 * 60];
for (String time : timePoints) {
String[] t = time.split(":");
int h = Integer.parseInt(t[0]);
int m = Integer.parseInt(t[1]);
if (mark[h * 60 + m]) return 0;
mark[h * 60 + m] = true;
}

int prev = 0, min = Integer.MAX_VALUE;
int first = Integer.MAX_VALUE, last = Integer.MIN_VALUE;
for (int i = 0; i < 24 * 60; i++) {
if (mark[i]) {
if (first != Integer.MAX_VALUE) {
min = Math.min(min, i - prev);
}
first = Math.min(first, i);
last = Math.max(last, i);
prev = i;
}
}

min = Math.min(min, (24 * 60 - last + first));

return min;
}
}

2017/03/28