日撸leetCode三道题---Day1---二分查找

二分查找时间复杂度为O(log(n))
针对有序数组

//定义查找区间
var low = 0
var high = n
//循环查找
while(low<high)//这里不能等于，不然会越界
{var mid = Math.floor(low+(high-low)/2)if(target>nums[mid]){low = mid+1}else if(target ==nums[mid]){return mid}else if(target<nums[mid]){high=mid}return low  //这是区间端口会相等，直接返回
}

菜鸟学算法第一天小结