35, 搜索插入位置

一、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 contains distinct values sorted in ascending order.

• -10⁴ <= target <= 10⁴

相关主题

• 数组
• 二分查找

二、题解

方法 1: 二分查找

pub fn search_insert(nums: Vec<i32>, target: i32) -> i32 {
    //Self::binary_search_1(nums, target)
    Self::binary_search_2(nums, target)
}
pub fn binary_search_1(nums: Vec<i32>, target: i32) -> i32 {
    let mut left = 0;
    let mut right = nums.len();
    while left < right {
        let mid = left + (right - left) / 2;
        if target < nums[mid] {
            right = mid;
        } else if nums[mid] < target {
            left = mid + 1;
        } else {
            return mid as i32;
        }
    }
    left as i32
}
pub fn binary_search_2(nums: Vec<i32>, target: i32) -> i32 {
    let mut left = 0_i32;
    let mut right = nums.len() as i32 - 1;
    while left <= right {
        let mid = left + (right - left) / 2;
        if target < nums[mid as usize] {
            right = mid - 1;
        } else if nums[mid as usize] < target {
            left = mid + 1;
        } else {
            return mid;
        }
    }
    left
}

public int searchInsert(int[] nums, int target) {
    //return this.binarySearch1(nums, target);
    return this.binarySearch2(nums, target);
}
public int binarySearch1(int[] nums, int target) {
    int left = 0;
    int right = nums.length;
    while (left < right) {
        int mid = left + (right - left) / 2;
        if (target < nums[mid]) {
            right = mid;
        } else if (nums[mid] < target) {
            left = mid + 1;
        } else {
            return mid;
        }
    }
    return left;
}
public int binarySearch2(int[] nums, int target) {
    int left = 0;
    int right = nums.length - 1;
    while (left <= right) {
        int mid = left + (right - left) / 2;
        if (target < nums[mid]) {
            right = mid - 1;
        } else if (nums[mid] < target) {
            left = mid + 1;
        } else {
            return mid;
        }
    }
    return left;
}