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35. 搜索插入位置

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。

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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

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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;
    }
}

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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;
    }
};

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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
}

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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
    }
}

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/**
 * @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;
};

方法二：二分查找（内置函数）

我们也可以直接使用内置函数进行二分查找。

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

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

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class Solution {
    public int searchInsert(int[] nums, int target) {
        int i = Arrays.binarySearch(nums, target);
        return i < 0 ? -i - 1 : i;
    }
}

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

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

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class Solution {
    /**
     * @param integer[] $nums
     * @param integer $target
     * @return integer
     */

    function searchInsert($nums, $target) {
        $key = array_search($target, $nums);
        if ($key !== false) {
            return $key;
        }

        $nums[] = $target;
        sort($nums);
        return array_search($target, $nums);
    }
}

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