34. Search for a Range（二分查找）

leetcode 34 Search for a Range

Given an array of integers sorted in ascending order, 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].

思路：

要求复杂度是logn，所以采用二分查找，先找到目标值，由于是有序的，再向左和向右查找找到起始和终止的index；

var searchRange = function(nums, target) {
    var res=[];
    var l=0;
    var r=nums.length-1;
    while(l<=r){
        var m=Math.floor((l+r)/2);
        if(nums[m]<target){
            l=m+1;
        }else if(nums[m]>target){
            r=m-1;
        }else{
            var left=m;
            var right=m;
            while(left>0 && nums[left-1]===nums[m]) --left;
            while(right<nums.length-1 && nums[right+1]===nums[m]) ++right;
            return [left,right];
        }
    }
    return [-1,-1];
};