LintCode 60: Search Insert Position

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博客围绕Search Insert Position问题展开，给定排序数组和目标值，若找到目标值则返回其索引，未找到则返回按顺序插入的索引，假设数组无重复元素，还给出示例，并提出O(log(n))时间复杂度挑战，最后给出二分查找解法代码。

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Search Insert Position
中文English
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.

Example
[1,3,5,6], 5 → 2

[1,3,5,6], 2 → 1

[1,3,5,6], 7 → 4

[1,3,5,6], 0 → 0

Challenge
O(log(n)) time

解法1：Binary Search
代码如下：

class Solution {
public:
    /**
     * @param A: an integer sorted array
     * @param target: an integer to be inserted
     * @return: An integer
     */
    int searchInsert(vector<int> &A, int target) {
        int n = A.size();
        if (n == 0) return 0;
        
        int start = 0, end = n - 1;
        while(start + 1 < end) {
            int mid = start + (end - start) / 2;
            if (A[mid] < target) {
                start = mid;
            } else if (A[mid] > target) {
                end = mid;
            } else {
                return mid;
            }
        }
        
        if (A[start] >= target) return start;
        if (A[end] >= target) return end;
        return end + 1;
    }
};