Question

35. 搜索插入位置

English Version

题目描述

给定一个排序数组和一个目标值，在数组中找到目标值，并返回其索引。如果目标值不存在于数组中，返回它将会被按顺序插入的位置。

请必须使用时间复杂度为 O(log n) 的算法。

 

示例 1:

输入: nums = [1,3,5,6], target = 5
输出: 2

示例 2:

输入: nums = [1,3,5,6], target = 2
输出: 1

示例 3:

输入: nums = [1,3,5,6], target = 7
输出: 4

 

提示:

• 1 <= nums.length <= 104
• -104 <= nums[i] <= 104
• nums 为 无重复元素 的 升序 排列数组
• -104 <= target <= 104

解法

方法一：二分查找

由于 $nums$ 数组已经有序，因此我们可以使用二分查找的方法找到目标值 $target$ 的插入位置。

时间复杂度 $O(\log n)$，空间复杂度 $O(1)$。其中 $n$ 为数组 $nums$ 的长度。

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

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

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

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

use std::cmp::Ordering;
impl Solution {
  pub fn search_insert(nums: Vec<i32>, target: i32) -> i32 {
    let mut left = 0;
    let mut right = nums.len();
    while left < right {
      let mid = left + (right - left) / 2;
      match nums[mid].cmp(&target) {
        Ordering::Less => {
          left = mid + 1;
        }
        Ordering::Greater => {
          right = mid;
        }
        Ordering::Equal => {
          return mid as i32;
        }
      }
    }
    left as i32
  }
}

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

方法二

class Solution:
  def searchInsert(self, nums: List[int], target: int) -> int:
    return bisect_left(nums, target)

class Solution {
public:
  int searchInsert(vector<int>& nums, int target) {
    return lower_bound(nums.begin(), nums.end(), target) - nums.begin();
  }
};

func searchInsert(nums []int, target int) int {
  return sort.SearchInts(nums, target)
}