pub fn has_duplicates(&self) -> bool {
        // Note: This is an O(n^2) algorithm, but requires no heap allocations. The benchmark
        // `bench_has_duplicates` in benches/message_processor.rs shows that this implementation is
        // ~50 times faster than using HashSet for very short slices.
        for i in 1..self.account_keys.len() {
            #[allow(clippy::integer_arithmetic)]
            if self.account_keys[i..].contains(&self.account_keys[i - 1]) {
                return true;
            }
        }
        false
    }
/// Return true if account_keys has any duplicate keys
Return true if account_keys has any duplicate keys
Return true if account_keys has any duplicate keys
pub fn has_duplicates(&self) -> bool {
        // Note: This is an O(n^2) algorithm, but requires no heap allocations. The benchmark
        // `bench_has_duplicates` in benches/message_processor.rs shows that this implementation is
        // ~50 times faster than using HashSet for very short slices.
        for i in 1..self.account_keys.len() {
            #[allow(clippy::integer_arithmetic)]
            if self.account_keys[i..].contains(&self.account_keys[i - 1]) {
                return true;
            }
        }
        false
    }
pub fn has_duplicates ( & self ) -> bool { // Note: This is an O(n^2) algorithm, but requires no heap allocations. The benchmark // `bench_has_duplicates` in benches/message_processor.rs shows that this implementation is // ~50 times faster than using HashSet for very short slices. for i in 1 .. self . account_keys . len ( ) { # [ allow ( clippy integer_arithmetic ) ] if self . account_keys [ i .. ] . contains ( & self . account_keys [ i - 1 ] ) { return true ; } } false }
Return true if account_keys has any duplicate keys
Return true if account_keys has any duplicate keys
&self