LeetCode – Search Insert Position-优快云博客

本文介绍了一种在已排序数组中查找目标值插入位置的算法，提供了两种解决方案：一种是简单的迭代方法，时间复杂度为O(n)；另一种是采用二分搜索的方法，将时间复杂度降低到O(log n)。通过具体示例展示了如何实现这两种算法。

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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

Solution 1

Naively, we can just iterate the array and compare target with ith and (i+1)th element. Time complexity is O(n)

public class Solution {
    public int searchInsert(int[] A, int target) {
 
        if(A==null) return 0;
 
        if(target <= A[0]) return 0;
 
        for(int i=0; i<A.length-1; i++){
            if(target > A[i] && target <= A[i+1]){
                return i+1;
            }
        }
 
        return A.length;
    }
}

Solution 2

This also looks like a binary search problem. We should try to make the complexity to be O(nlogn).

public class Solution {
    public int searchInsert(int[] A, int target) {
        if(A==null||A.length==0)
            return 0;
 
        return searchInsert(A,target,0,A.length-1);
    }
 
    public int searchInsert(int[] A, int target, int start, int end){
        int mid=(start+end)/2;
 
        if(target==A[mid]) 
            return mid;
        else if(target<A[mid]) 
            return start<mid?searchInsert(A,target,start,mid-1):start;
        else 
            return end>mid?searchInsert(A,target,mid+1,end):(end+1);
    }
}