35. Search Insert Position

搜索插入位置算法解析

最新推荐文章于 2025-11-26 15:16:48 发布

转载最新推荐文章于 2025-11-26 15:16:48 发布 · 69 阅读

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#数据结构与算法 #java

本文详细解析了在已排序数组中查找目标值并确定其插入位置的经典二分查找算法。该算法可在未找到目标值时返回正确的插入位置，并且具有O(logn)的时间复杂度。

题目：

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.

Here are few examples.
[1,3,5,6], 5 → 2
[1,3,5,6], 2 → 1
[1,3,5,6], 7 → 4
[1,3,5,6], 0 → 0

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Array Binary Search

链接： http://leetcode.com/problems/search-insert-position/

题解：

在给定数组中找到target，假如找到返回所在index，假如不在数组中返回应该插入的位置。典型的binary search，假如目标不在数组中，则最后循环结束时left所在index为应该插入的位置，right所在index比left小1。

Time Complexity - O(logn)， Space Complexity - O(1)。

public class Solution {
    public int searchInsert(int[] nums, int target) {   //binary search
        if(nums == null || nums.length == 0)
            return 0;
        int left = 0, right = nums.length - 1;
        
        while(left <= right){
            int mid = left + (right - left) / 2;
            if(nums[mid] == target)
                return mid;
            else if(nums[mid] < target)
                left++;
            else
                right--;
        }
        
        return left;
    }
}

二刷：

也是binary search，最后返回lo就可以了。

Java:

public class Solution {
    public int searchInsert(int[] nums, int target) {
        if (nums == null || nums.length == 0) {
            return -1;
        }
        int lo = 0, hi = nums.length - 1;
        while (lo <= hi) {
            int mid = lo + (hi - lo) / 2;
            if (target == nums[mid]) {
                return mid;
            } else if (target < nums[mid]) {
                hi = mid - 1;
            } else {
                lo = mid + 1;
            }
        }
        return lo;
    }
}

三刷:

Java:

public class Solution {
    public int searchInsert(int[] nums, int target) {
        if (nums == null || nums.length == 0) return 0;
        int lo = 0, hi = nums.length - 1;
        
        while (lo <= hi) {
            int mid = lo + (hi - lo) / 2;
            if (nums[mid] < target) lo = mid + 1;
            else if (nums[mid] > target) hi = mid - 1;
            else return mid;
        }
        return lo;    
    }
}