35.Search Insert Position

tags: `Easy`,`Array`,`Binary Search`

題目描述

Given a sorted array of distinct integers and a target value, return the index if the target is found. If not, return the index where it would be if it were inserted in order.

You must write an algorithm with O(log n) runtime complexity.

範例

Example 1:

Input: nums = [1,3,5,6], target = 5
Output: 2

Example 2:

Input: nums = [1,3,5,6], target = 2
Output: 1

Example 3:

Input: nums = [1,3,5,6], target = 7
Output: 4

Constraints:

1 <= nums.length <= 10⁴
-10⁴ <= nums[i] <= 10⁴
nums contains distinct values sorted in ascending order.
-10⁴ <= target <= 10⁴

解答

Python












class Solution:
    def searchInsert(self, nums: List[int], target: int) -> int:
        L, R = 0, len(nums) - 1
        while L <= R:
            M = (L + R) // 2
            if nums[M] == target:
                return M
            if target < nums[M]:
                R = M - 1
            else:
                L = M + 1
        return R + 1

Yen-Chi ChenMon, Feb 20, 2023



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

Yen-Chi ChenMon, Feb 20, 2023












class Solution:
    def searchInsert(self, nums: List[int], target: int) -> int:
        l, r = 0, len(nums)

        while l < r:
            mid = l + (r - l) // 2
            if nums[mid] >= target:
                r = mid
            else:
                l = mid + 1

        return l

Ron ChenWed, Feb 22, 2023

Javascript















function searchInsert(nums, target) {
  let max = nums.length - 1;
  let min = 0;
  while (min <= max) {
    const position = ~~((max + min) / 2);
    if (nums[position] === target) return position;
    if (nums[position] > target) {
      max = position - 1;
    } else {
      min = position + 1;
    }
  }

  return max + 1;
}

MasrgoatMon, Feb 20, 2023

Reference

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