Question

704. 二分查找

English Version

题目描述

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  ，写一个函数搜索 nums 中的 target，如果目标值存在返回下标，否则返回 -1。

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

 

提示：

1. 你可以假设 nums 中的所有元素是不重复的。
2. n 将在 [1, 10000]之间。
3. nums 的每个元素都将在 [-9999, 9999]之间。

解法

方法一

class Solution:
  def search(self, nums: List[int], target: int) -> int:
    left, right = 0, len(nums) - 1
    while left < right:
      mid = (left + right) >> 1
      if nums[mid] >= target:
        right = mid
      else:
        left = mid + 1
    return left if nums[left] == target else -1

class Solution {
  public int search(int[] nums, int target) {
    int left = 0, right = nums.length - 1;
    while (left < right) {
      int mid = (left + right) >> 1;
      if (nums[mid] >= target) {
        right = mid;
      } else {
        left = mid + 1;
      }
    }
    return nums[left] == target ? left : -1;
  }
}

class Solution {
public:
  int search(vector<int>& nums, int target) {
    int left = 0, right = nums.size() - 1;
    while (left < right) {
      int mid = left + right >> 1;
      if (nums[mid] >= target)
        right = mid;
      else
        left = mid + 1;
    }
    return nums[left] == target ? left : -1;
  }
};

func search(nums []int, target int) int {
  left, right := 0, len(nums)-1
  for left < right {
    mid := (left + right) >> 1
    if nums[mid] >= target {
      right = mid
    } else {
      left = mid + 1
    }
  }
  if nums[left] == target {
    return left
  }
  return -1
}

use std::cmp::Ordering;

impl Solution {
  pub fn search(nums: Vec<i32>, target: i32) -> i32 {
    let mut l = 0;
    let mut r = nums.len();
    while l < r {
      let mid = (l + r) >> 1;
      match nums[mid].cmp(&target) {
        Ordering::Less => {
          l = mid + 1;
        }
        Ordering::Greater => {
          r = mid;
        }
        Ordering::Equal => {
          return mid as i32;
        }
      }
    }
    -1
  }
}

/**
 * @param {number[]} nums
 * @param {number} target
 * @return {number}
 */
var search = function (nums, target) {
  let left = 0;
  let right = nums.length - 1;
  while (left < right) {
    const mid = (left + right) >> 1;
    if (nums[mid] >= target) {
      right = mid;
    } else {
      left = mid + 1;
    }
  }
  return nums[left] == target ? left : -1;
};

方法二

use std::cmp::Ordering;

impl Solution {
  fn binary_search(nums: Vec<i32>, target: i32, l: usize, r: usize) -> i32 {
    if l == r {
      return if nums[l] == target { l as i32 } else { -1 };
    }
    let mid = (l + r) >> 1;
    match nums[mid].cmp(&target) {
      Ordering::Less => Self::binary_search(nums, target, mid + 1, r),
      Ordering::Greater => Self::binary_search(nums, target, l, mid),
      Ordering::Equal => mid as i32,
    }
  }

  pub fn search(nums: Vec<i32>, target: i32) -> i32 {
    let r = nums.len() - 1;
    Self::binary_search(nums, target, 0, r)
  }
}