Search Insert Position leetcode java

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本文介绍了一道二分查找变种题目的解决方法，该题要求在有序数组中查找目标值的位置，若不存在则返回应插入的位置。通过详细步骤解析及代码实现，帮助读者理解这一算法的应用。

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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

本题是基本考察二分查找的题目，与基本二分查找方法不同的地方是，二分查找法当查找的target不在list中存在时返回-1，而本题则需要返回该target应在此list中插入的位置。

当循环结束时，如果没有找到target，那么low一定停target应该插入的位置上，high一定停在恰好比target小的index上。

[1,3,5,6], 7

low = 0, high = 3

step1: mid = 1

A[mid] = 3, 3<7

low = mid + 1 = 2

low = 2, high = 3

step2: mid = 2

A[mid] = 5, 5<7

low = mid + 1 = 3

low = 3, high = 3

step3: mid = 3

A[mid] = 6, 6<7

low = mid + 1 = 4

low = 4, high = 3

return low = 4;

[1,3,5,6], 0

low = 0, high = 3

step1: mid = 1

A[mid] = 3, 3>0

high = mid - 1 = 0

low = 0, high = 0

step2: mid = 0

A[mid] = 1, 1>0

high = mid - 1 = -1

low = 0, high = -1

return 0

public int searchInsert(int[] A, int target) {
        if(A == null||A.length == 0)
            return 0;
        int low = 0, high = A.length-1;
        
        while(low <= high){
            int mid = (low + high) / 2;
            
            if(A[mid] > target)
                high = mid - 1;
            else if(A[mid] < target)
                low = mid + 1;
            else
                return mid;
        }
        
        return low;
    }

Reference: http://blog.youkuaiyun.com/linhuanmars/article/details/31354941