leetcode35 Search Insert Position

# 35. 搜索插入位置 (opens new window)

# 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。

你可以假设数组中无重复元素。

示例 1:

输入: [1,3,5,6], 5
输出: 2

示例 2:

输入: [1,3,5,6], 2
输出: 1

示例 3:

输入: [1,3,5,6], 7
输出: 4

示例 4:

输入: [1,3,5,6], 0
输出: 0

# 解法

二分查找。

# Python3

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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# Java

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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# C++

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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# Go

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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# JavaScript

/**
 * @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;
};

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# ...

编辑

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上次更新: 2021/10/30, 12:58:38

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