[leetcode javascript解題]Search for a range

leetcode 第34題”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]

.

這首先條件是給了我們一個從小到大已經排列好的數組，然後找到數組中值等于value的起始鍵和終止鍵，由于題目說複雜度控制在O(logn)其實就已經變相暗示用二分法解決了，最傳統的方法當然是雙指針，但是自己想練練自己遞歸解題的思路，就用遞歸解的：

/**
 * @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;
};