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LeetCode-393-UTF-8-Validation.java
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LeetCode-393-UTF-8-Validation.java
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/*
https://www.youtube.com/watch?v=-r1cL8lCLeM
https://leetcode.com/problems/utf-8-validation/discuss/87464/Bit-Manipulation-Java-6ms
Traverse the given int array,
Find the number of bytes of that number,
check the next n - 1 bytes start with 10, if not return false;
Finally, return true.
Analysis:
197
7 6 5 4 3 2 1 0
128 64 32 16 8 4 2 1
1 1 0 0 0 1 0 1
197 = 11000101
1 byte = 8 bits, so the range of 1 byte is [0, 255]
*/
class Solution {
public boolean validUtf8(int[] data) {
int count = 0; // count means the number of bytes that are expected to start with "10"
for (int d : data) {
if (count == 0) {
// means d is the first character, or the invalid case, like [110xxxxx, 10xxxxxx, 10xxxxxx]
if ((d >> 5) == 0b110) {
count = 1;
} else if ((d >> 4) == 0b1110) {
count = 2;
} else if ((d >> 3) == 0b11110) {
count = 3;
} else if ((d >> 7) != 0b0) {
// b should be the case [0xxxxxxx], if it doesn't start with "0", then it's wrong.
return false;
}
} else {
// means d is not the first byte, we check current d if it starts with "10"
if ((d >> 6) != 0b10) return false;
else count--;
}
}
return count == 0;
}
}