list1.Add(new { id = "1", num = 11, time = "2015-10-10 10:10:10" });
list1.Add(new { id = "2", num = 22, time = "2015-10-10 11:11:11" });
list1.Add(new { id = "3", num = 33, time = "2015-10-10 12:12:12" });
List<dynamic> list2 = new List<dynamic>();
list2.Add(new { id = "11", num = 18, time = "2015-10-10 10:10:10" });
list2.Add(new { id = "21", num = 10, time = "2015-10-10 11:11:11" });
list2.Add(new { id = "3", num = 55, time = "2015-10-10 12:12:12" });
有上面2个数组,以两个数组的时间相等,判断出2个num值偏差比较大的那条数据,如上列2列要返回 list1.Add(new { id = "3", num = 33, time = "2015-10-10 12:12:12" });
linq的方法如何实现,或者有其他更好的方法也可以。