leetcode-35-search insert position

Given a sorted array and a target value, return the index if the target is found. If not, return the index where it would be if it were inserted in order.

You may assume no duplicates in the array.

Here are few examples.
[1,3,5,6], 5 → 2
[1,3,5,6], 2 → 1
[1,3,5,6], 7 → 4
[1,3,5,6], 0 → 0

大意：在有序数组中插入元素。

思路：二分查找，找到target就返回此刻索引。否则维护和target距离最小的那个索引。最后判断该位置元素和target大小比较。

python一遍AC。

class Solution(object):    def searchInsert(self, nums, target):        """        :type nums: List[int]        :type target: int        :rtype: int        """        l,r=0,len(nums)-1        ans=0        dismin=9999        if r==-1:return 0        while(l<=r):            m=(l+r)/2            if nums[m]==target:                return m            if nums[m]<target:                dis=abs(nums[m]-target)                if dis<dismin:                    dismin=dis                    a=m                l=m+1            else:                dis=abs(nums[m]-target)                if dis<dismin:                    dismin=dis                    a=m                r=m-1        if nums[a]<target:            return a+1        else:            return a

再一考虑，发现不需要维护dismin。直接二分查找。最后返回l and r。
找到了返回m，找不到返回left即可。

class Solution(object):    def searchInsert(self, nums, target):        """        :type nums: List[int]        :type target: int        :rtype: int        """        l,r=0,len(nums)-1        ans=0        dismin=9999        if r==-1:return 0        while(l<=r):            m=(l+r)/2            if nums[m]==target:                return m            if nums[m]<target:                l=m+1            else:                r=m-1        return l

c++:

class Solution {public:    int searchInsert(vector<int>& nums, int target)      {          int n = nums.size();          int low = 0;          int high = n -1;          int mid;          while(low <= high)          {               mid = low + (high - low)/2;              if(target == nums[mid])                  return mid;              else if(target < nums[mid])              high = mid - 1;              else                   low = mid + 1;          }          return low;      } };