注意两点:
1、计算两个时刻的时间差时,将两个时刻都换算成分钟(h*60+m),计算差的绝对值diff。若diff > 12*60,diff应该变为24*60-diff
2、注意不要将所有时刻的差值都存储下来,因为需要的空间为O(n*n),这样会报内存错误。
class Solution { public: int findMinDifference(vector<string>& timePoints) { int n = timePoints.size(); int min = 720, k = 0; vector<int> minutePoints(n); for(int i = 0; i < n; ++ i){ string s = timePoints[i]; minutePoints[i] = ((s[0]-'0')*10+(s[1]-'0'))*60 + ((s[3]-'0')*10+(s[4]-'0')); } for(int i = 0; i < n; ++ i){ for(int j = i+1; j < n; ++ j){ int diff = abs(minutePoints[i]-minutePoints[j]); int actualDiff = diff < 720 ? diff : (1440 - diff); if(actualDiff < min){ min = actualDiff; } } } return min; } };