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14. 最长公共前缀
编写一个函数来查找字符串数组中的最长公共前缀。
如果不存在公共前缀，返回空字符串 “”。
方法一 竖向扫描法
个人感觉纵向扫描方式比较直观，符合人类理解方式，从前往后遍历所有字符串的每一列，比较相同列上的字符是否相同，如果相同则继续对下一列进行比较，如果不相同则当前列不再属于公共前缀，当前列之前的部分为最长公共前缀。
Swift
func longestCommonPrefix(_ strs: [String]) -> String {guard let firstStr = strs.first, !firstStr.isEmpty else { return "" }for i in 0..<firstStr.count {for j in 1..<strs.count {if strs[j].count == i || strs[j][strs[j].index(strs[j].startIndex, offsetBy: i)] != firstStr[firstStr.index(firstStr.startIndex, offsetBy: i)] {return String(firstStr.prefix(i))}}}return firstStr}
OC
-(NSString *)longestCommonPrefix:(NSArray <NSString *>*)strs {if (strs.count <= 0) {return @"";}NSString *firstStr = strs.firstObject;NSInteger len = firstStr.length;for (NSInteger i=0; i<len; i++) {for (NSInteger j=1; j<strs.count; j++) {if (strs[j].length == i || [strs[j] characterAtIndex:i] != [firstStr characterAtIndex:i]) {return [firstStr substringToIndex:i];}}}return firstStr;
}
方法二 有序首尾比较法
有序首尾比较法，先对数组进行排序，巧妙利用排序后的顺序及值之间的关系，只比较首尾两个字符串即可。
Swift
func longestCommonPrefix(_ strs: [String]) -> String {let strs = strs.sorted()let start = strs.first!let end = strs.last!var res = ""for i in 0..<start.count {let s = start[start.index(start.startIndex, offsetBy: i)]if s == end[end.index(end.startIndex, offsetBy: i)]{res.append(s)}else {break}}return res}
OC
//有序首尾比较法
-(NSString *)longestCommonPrefix:(NSArray *)strs {NSArray *sortedStrs = [strs sortedArrayUsingComparator:^NSComparisonResult(NSString *obj1, id _Nonnull obj2) {return [obj1 compare:obj2 options:NSCaseInsensitiveSearch];}];NSString *res = @"";NSString *firstStr = sortedStrs.firstObject;NSString *lastStr = sortedStrs.lastObject;for (NSInteger i=0; i<firstStr.length; i++) {if ([firstStr characterAtIndex:i] == [lastStr characterAtIndex:i]) {unichar c = [firstStr characterAtIndex:i];res = [res stringByAppendingString:[NSString stringWithCharacters:&c length:1]];}else {break;}}return res;
}