[LeetCode35]Search Insert Position

最新推荐文章于 2025-11-24 14:44:53 发布

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#数组 #leetcode #binary search

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本文介绍了一个算法问题：在已排序的数组中查找目标值的插入位置。提供了详细的分析过程及Java和C++实现代码，利用二分搜索快速定位元素。

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

Analysis:

could use binary search, there has a additional condition if(mid>left && A[mid]>target && A[mid-1]<target)

Java

public int searchInsert(int[] A, int target) {
        // IMPORTANT: Please reset any member data you declared, as
        // the same Solution instance will be reused for each test case.
        int low = 0;
		int high = A.length-1;
		if(target < A[0])
        	return 0;
		if(target > A[A.length-1])
        	return A.length;
		while(low <= high){
			int mid = (low+high)/2;
			if(A[mid]==target)
				return mid;
			if(mid>1 && A[mid]>target && A[mid-1]<target) 
				return mid;
			if(A[mid]>target)
				high = mid-1;
			else
				low = mid+1;
		}
		return 1;
    }

update

public int searchInsert(int[] nums, int target) {
        int len = nums.length;
        if(len<=0) return 0;
        if(target<=nums[0]) return 0;
        if(target>nums[len-1]) return len;
        int left = 0;
        int right = len-1;
        while(left<=right) {
            int mid = left + (right-left)/2;
            if(nums[mid] == target) return mid;
            if(nums[mid] > target) {
                if(target>nums[mid-1]) return mid;
                else right = mid-1;
            }else {
                if(target<nums[mid+1]) return mid+1;
                else left = mid+1;
            }
        }
        return left;
    }

c++

int searchInsert(int A[], int n, int target) {
       if(A == NULL|| target<A[0]) return 0;
    if(target>A[n-1]) return n;
    int l = 0;
    int r = n-1;
    while(l<=r){
        int m = (l+r)/2;
        if(A[m]==target) return m;
        if(m>l&& A[m]>target && A[m-1]<target) return m;
        if(A[m]>target)
            r = m-1;
        else
            l = m+1;
    }
    return l; 
    }