README.md

Search Insert Position

Description

Given a sorted array 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 may assume no duplicates in the array.

Example 1:

Input: [1,3,5,6], 5
Output: 2

Example 2:

Input: [1,3,5,6], 2
Output: 1

Example 3:

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

Example 1:

Input: [1,3,5,6], 0
Output: 0

Tags: Array, Binary Search

思路

题意是让你从一个没有重复元素的已排序数组中找到插入位置的索引。因为数组已排序，所以我们可以想到二分查找法，因为查找到的条件是找到第一个等于或者大于 target 的元素的位置，所以二分法略作变动即可。

Java:

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

kotlin(216ms/90.00%):

class Solution {
    fun searchInsert(nums: IntArray, target: Int): Int {
        var left = 0
        var right = nums.size - 1
        while (left <= right) {
            val mid = (left + right) / 2
            when {
                nums[mid] == target -> return mid
                nums[mid] < target -> left = mid + 1
                else -> right = mid - 1
            }
        }
        return left
    }
}

javascript

var searchInsert = function(nums, target) {
    if (nums.length === 1) {
        return nums[0] >= target?0:1
    }
    for (let i = 0; i < nums.length - 1; i++) {
        if (nums[i] === target) {
            return i
        }
        if (nums[i] < target && target <= nums[i + 1]) {
            return i+1
        }
        if (nums[i] > target) {
            return 0
        }
    }
    return nums.length
};

结语

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