704. 二分查找 ★
力扣题目链接,给定一个 n 个元素有序的(升序)整型数组 nums 和一个目标值 target ,搜索 nums 中的 target,如果存在返回下标,否则返回 -1。n 将在 [1, 10000]之间。
- 可以假设 nums中的所有元素是不重复的。
- n将在- [1, 10000]之间。
- nums的每个元素都将在- [-9999, 9999]之间。
示例 1:
输入: nums = [-1,0,3,5,9,12], target = 9
输出: 4
解释: 9 出现在 nums 中并且下标为 4示例 2:
输入: nums = [-1,0,3,5,9,12], target = 2
输出: -1
解释: 2 不存在 nums 中因此返回 -1本地练习
            
            
              rust
              
              
            
          
          pub struct Solution;
use std::cmp::Ordering;
impl Solution {
    pub fn search(nums: Vec<i32>, target: i32) -> i32 {
        
    }
}
fn main() {
    let res = [(vec![-1, 0, 3, 5, 9, 12], 9), (vec![-1, 0, 3, 5, 9, 12], 2)]
        .iter().map(|x| Solution::search(x.0.to_vec(), x.1)).collect::<Vec<_>>();
    println!("{:?}: {:?}", vec![4, -1] == res, res);
}Rust答案
- (版本一)左闭右开区间
            
            
              rust
              
              
            
          
          use std::cmp::Ordering;
impl Solution {
    pub fn search(nums: Vec<i32>, target: i32) -> i32 {
        let (mut left, mut right) = (0, nums.len());
        while left < right {
            let mid = (left + right) / 2;
            match nums[mid].cmp(&target) {
                Ordering::Less => left = mid + 1,
                Ordering::Greater => right = mid,
                Ordering::Equal => return mid as i32,
            }
        }
        -1
    }
}- (版本二)左闭右闭区间
            
            
              rust
              
              
            
          
          use std::cmp::Ordering;
impl Solution {
    pub fn search(nums: Vec<i32>, target: i32) -> i32 {
        let (mut left, mut right) = (0, nums.len());
        while left <= right {
            let mid = (right + left) / 2;
            match nums[mid].cmp(&target) {
                Ordering::Less => left = mid + 1,
                Ordering::Greater => right = mid - 1,
                Ordering::Equal => return mid as i32,
            }
        }
        -1
    }
}