Search Insert Position [LEETCODE]

最新推荐文章于 2025-09-12 18:45:12 发布

weixin_33910460

最新推荐文章于 2025-09-12 18:45:12 发布

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CC 4.0 BY-SA版权

文章标签： c/c++

原文链接：http://www.cnblogs.com/scenix/p/3342464.html

本文介绍了一个简单的二分查找算法的应用实例，该实例用于在一个已排序的数组中查找目标值的位置或者确定插入位置。首先实现了一个O(n)的简单查找方法，然后通过二分查找将时间复杂度降低到O(log n)，并提供了详细的代码实现。

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

==================================================================================================

First , code a O(n) solution, see if it exceeds time limit:

class Solution {
public:
    int searchInsert(int A[], int n, int target) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        int i = 0;
        for(; i < n; i++) {
            if(target <= A[i]) {
                return i;
            }
        }
        return i;
    }
};

Submission Result: Accepted

OK, this problem does not seem have much expectation in me...

But I still want to challenge myself with it.

So I tried a O(lgn) solution using binary search.

class Solution {
public:
    int searchInsert(int A[], int n, int target) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        int left = 0;
        int right = n - 1;
        int mid = (left + right) / 2;
        while(1){
            if(target == A[mid]) {
                return mid;    
            }
            if(target > A[mid]){
                //find in right sector
                left = mid + 1;
            }
            else if(target < A[mid]){
                //find in left sector
                right = mid - 1;
            }
            if(left > right){
                if(target > A[mid]){
                    return mid + 1;
                } else {
                    return mid;
                }
            }
            mid = (left + right) / 2;
        }
    }
};