題目要求

A character in UTF8 can be from 1 to 4 bytes long, subjected to the following rules:For 1-byte character, the first bit is a 0, followed by its unicode code. For n-bytes character, the first n-bits are all one's, the n+1 bit is 0, followed by n-1 bytes with most significant 2 bits being 10. This is how the UTF-8 encoding would work:Char. number range | UTF-8 octet sequence(hexadecimal) | (binary)--------------------+---------------------------------------------0000 0000-0000 007F | 0xxxxxxx0000 0080-0000 07FF | 110xxxxx 10xxxxxx0000 0800-0000 FFFF | 1110xxxx 10xxxxxx 10xxxxxx0001 0000-0010 FFFF | 11110xxx 10xxxxxx 10xxxxxx 10xxxxxx Given an array of integers representing the data, return whether it is a valid utf-8 encoding.Note: The input is an array of integers. Only the least significant 8 bits of each integer is used to store the data. This means each integer represents only 1 byte of data.Example 1:data = [197, 130, 1], which represents the octet sequence: 11000101 10000010 00000001.Return true. It is a valid utf-8 encoding for a 2-bytes character followed by a 1-byte character. Example 2:data = [235, 140, 4], which represented the octet sequence: 11101011 10001100 00000100.Return false. The first 3 bits are all one's and the 4th bit is 0 means it is a 3-bytes character. The next byte is a continuation byte which starts with 10 and that's correct. But the second continuation byte does not start with 10, so it is invalid.

檢驗整數數組能否構成合法的UTF8編碼的序列。UTF8的字節編碼規則如下：

每個UTF8字符包含1~4個字節

如果只包含1個字節，則該字節以0作為開頭，剩下的位隨意

如果包含兩個或兩個以上字節，則起始字節以n個1和1個0開頭，例如，如果該UTF8字符包含兩個字節，則第一個字節以110開頭，同理，三個字符的第一個字節以1110開頭。剩余的字節必須以10開頭。

思路和代碼

首先我們整理一下，每一種類型的UTF8字符包含什么樣的規格：

只包含一個字節，該字節格式為0xxxxxxx，則轉換為整數的話，該整數必須小于128（1000000）

包含多個字節，則頭字節格式為110xxxxx, 1110xxxx, 11110xxx。而緊跟其后的字符必須格式為10xxxxxx。

綜上所述：

num<1000000: 單字節

10000000=<num<11000000: 多字節字符的跟隨字節

11000000<=num<11100000: 兩個字節的起始字節

11100000<=num<11110000: 三個字節的起始字節

11110000<=num<11111000: 四個字節的起始字節

下面分別是這題的兩種實現：

遞歸實現：

private static final int ONE_BYTE = 128; //10000000private static final int FOLLOW_BYTE = 192; //11000000private static final int TWO_BYTE = 224; //11100000private static final int THREE_BYTE = 240;//11110000private static final int FOUR_BYTE = 248;//11111000public boolean validUtf8(int[] data) {return validUtf8(data, 0);}public boolean validUtf8(int[] data, int startAt) {if(startAt >= data.length) return true;int first = data[startAt];int followLength = 0;if(first < ONE_BYTE) {return validUtf8(data, startAt+1);}else if(first < FOLLOW_BYTE){return false;}else if(first <TWO_BYTE) {followLength = 2;}else if(first < THREE_BYTE) {followLength = 3;}else if(first < FOUR_BYTE) {followLength = 4;}else {return false;}if(startAt + followLength > data.length) return false; for(int i = 1 ; i<followLength ; i++) {int next = data[startAt + i];if(next < ONE_BYTE || next >= FOLLOW_BYTE) {return false;}}return validUtf8(data, startAt + followLength);}

循環實現：

private static final int ONE_BYTE = 128; //10000000private static final int FOLLOW_BYTE = 192; //11000000private static final int TWO_BYTE = 224; //11100000private static final int THREE_BYTE = 240;//11110000private static final int FOUR_BYTE = 248;//11111000public boolean validUtf8(int[] data) {return validUtf8(data, 0);}public boolean validUtf8(int[] data, int startAt) {int followCount = 0;for(int num : data) {if(num < ONE_BYTE) {if(followCount != 0) {return false;}}else if(num < FOLLOW_BYTE) {if(followCount == 0) {return false;}followCount--;}else if(num < TWO_BYTE) {if(followCount != 0) {return false;}followCount = 1;}else if(num < THREE_BYTE) {if(followCount != 0) {return false;}followCount = 2;}else if(num < FOUR_BYTE) {if(followCount != 0) {return false;}followCount = 3;}else {return false;}}return followCount == 0;} 創作挑戰賽新人創作獎勵來咯，堅持創作打卡瓜分現金大獎

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