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/// 1. Two Sum (Easy)
///
/// Given an array of integers nums and an integer target, return indices of the
/// two numbers such that they add up to target.
///
/// You may assume that each input would have exactly one solution, and you may
/// not use the same element twice.
///
/// You can return the answer in any order.
///
/// Example 1:
///
/// Input: nums = [2,7,11,15], target = 9
/// Output: [0,1]
/// Explanation: Because nums[0] + nums[1] == 9, we return [0, 1].
///
/// Example 2:
///
/// Input: nums = [3,2,4], target = 6
/// Output: [1,2]
///
/// Example 3:
///
/// Input: nums = [3,3], target = 6
/// Output: [0,1]
///
/// Constraints:
///
/// * 2 <= nums.length <= 104
/// * -109 <= nums[i] <= 109
/// * -109 <= target <= 109
/// * Only one valid answer exists.
///
/// Follow-up: Can you come up with an algorithm that is less than O(n2) time
/// complexity?
use leetcode::*;
struct Solution {}
impl Solution {
pub fn two_sum(nums: Vec<i32>, target: i32) -> Vec<i32> {
let mut nums: Vec<_> = nums.iter().enumerate().collect();
nums.sort_by(|(_ai, a), (_bi, b)| a.cmp(b));
let mut i = 0usize;
let mut k = nums.len() - 1;
let mut result = Vec::new();
loop {
if nums[i].1 + nums[k].1 == target {
break;
} else if nums[i + 1].1 + nums[k].1 <= target {
i += 1;
} else {
k -= 1;
}
}
result.push(nums[i].0 as i32);
result.push(nums[k].0 as i32);
result
}
}
pub fn main() {
println!("{:?}", Solution::two_sum(arg_into(1), arg_into(2)))
}
#[cfg(test)]
mod tests {
use crate::Solution;
use leetcode::vi;
#[test]
fn test() {
assert_eq!(Solution::two_sum(vi("[2,7,11,15]"), 9), vi("[0,1]"));
assert_eq!(Solution::two_sum(vi("[3,2,4]"), 6), vi("[1,2]"));
assert_eq!(Solution::two_sum(vi("[3,3]"), 6), vi("[0,1]"));
}
}
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