LintCode 459: Closest Number in Sorted Array

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博客围绕在升序排序的整数数组中查找最接近给定目标值的元素索引展开，给出多个示例展示不同输入下的输出结果，还提出了O(logn)时间复杂度的挑战，最后给出了二分法的解法。

Closest Number in Sorted Array
中文English
Given a target number and an integer array A sorted in ascending order, find the index i in A such that A[i] is closest to the given target.

Return -1 if there is no element in the array.

Example
Example 1:

Input: [1, 2, 3] , target = 2
Output: 1.
Example 2:

Input: [1, 4, 6], target = 3
Output: 1.
Example 3:

Input: [1, 4, 6], target = 5,
Output: 1 or 2.
Challenge
O(logn) time complexity.

Notice
There can be duplicate elements in the array, and we can return any of the indices with same value.

解法1：二分法。
代码如下：

class Solution {
public:
    /**
     * @param A: an integer array sorted in ascending order
     * @param target: An integer
     * @return: an integer
     */
    int closestNumber(vector<int> &A, int target) {
        int n = A.size();
        if (n == 0) return 0;
        if (n == 1) return A[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 (abs(target - A[start]) < abs(target - A[end])) {
            return start;
        } else {
            return end;
        }
    }
};