Solve first-missing-positive.
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@ -0,0 +1,36 @@
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pub fn first_missing_positive(mut numbers: Vec<i32>) -> i32 {
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let numbers_length = numbers.len();
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// Mark all numbers below 1 and above the maximum range as i32::MAX.
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for index in 0..numbers_length {
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if numbers[index] < 1 || numbers[index] > numbers_length as i32 {
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numbers[index] = i32::MAX;
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}
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}
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for index in 0..numbers_length {
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// For each index get its number and subtract 1.
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let index = numbers[index].abs() as usize - 1;
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// If that index then isn't i32::MAX - 1 (ie. negative or above range).
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if index as i32 != i32::MAX - 1 {
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// Then mark the number for that index as negative.
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numbers[index] = -numbers[index].abs();
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}
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}
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// Then the first positive number we find, the index of that number will be
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// the first missing positive. And if there are only negative numbers, then
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// the next positive number is the total amount of numbers + 1.
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numbers
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.into_iter()
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.enumerate()
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.find_map(|(index, number)| {
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if number.is_positive() {
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Some(index + 1)
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} else {
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None
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}
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})
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.unwrap_or_else(|| numbers_length + 1) as i32
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}
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@ -3,6 +3,7 @@ pub mod contains_duplicate;
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pub mod contains_duplicate_ii;
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pub mod excel_sheet_column_number;
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pub mod excel_sheet_column_title;
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pub mod first_missing_positive;
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pub mod fizz_buzz;
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pub mod implement_strstr;
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pub mod integer_to_roman;
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@ -0,0 +1,14 @@
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use leetcode::first_missing_positive::first_missing_positive;
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use test_case::test_case;
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#[test_case(&[1, 2, 0], 3; "example 1")]
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#[test_case(&[3, 4, -1, 1], 2; "example 2")]
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#[test_case(&[7, 8, 9, 11, 12], 1; "example 3")]
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#[test_case(&[1, 3, 2], 4; "maximum")]
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#[test_case(&[1, 3, 3], 2; "duplicate")]
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#[test_case(&[0], 1; "zero")]
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#[test_case(&(0..=500000).into_iter().collect::<Vec<i32>>(), 500001; "massive")]
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fn test_first_missing_positive(input: &[i32], expected: i32) {
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assert_eq!(first_missing_positive(input.to_vec()), expected);
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}
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