[leetcode javascript解题]Search for a range

leetcode 第34题”Search for a range”描述如下：

Given a sorted array of integers, find the starting and ending position of a given target value.

Your algorithm’s runtime complexity must be in the order of O(log n).

If the target is not found in the array, return

[-1, -1]

.

For example,

Given

[5, 7, 7, 8, 8, 10]

and target value

8

,

return

[3, 4]

.

这首先条件是给了我们一个从小到大已经排列好的数组，然后找到数组中值等于value的起始键和终止键，由于题目说复杂度控制在O(logn)其实就已经变相暗示用二分法解决了，最传统的方法当然是双指针，但是自己想练练自己递归解题的思路，就用递归解的：

/**
 * @param {number[]} nums
 * @param {number} target
 * @return {number[]}
 */
var search = function(left, right, result, nums, target) {
    if (left > right) {
        return;
    }
    var pos = Math.floor((right + left) / );
    if (nums[pos] === target) {
        if (result[] === - && result[] === -) {
            result[] = result[] = pos;
        } else {
            result[] = pos < result[] ? pos : result[];
            result[] = pos > result[] ? pos : result[];
        }
        search(left, pos - , result, nums, target);
        search(pos + , right, result, nums, target);
    } else if (nums[pos] > target) {
        search(left, pos - , result, nums, target);
    } else {
        search(pos + , right, result, nums, target);
    }
};
var searchRange = function(nums, target) {
    var result = [-, -];
    search(, nums.length - , result, nums, target);
    return result;
};