算法第三周

4. Median of Two Sorted Arrays

There are two sorted arrays nums1 and nums2 of size m and n respectively.Find the median of the two sorted arrays. The overall run time complexity should be O(log (m+n)).

目标：在两个已排序好的数组中找到中间值。

遍历两个数组，每次将当前最小值压入向量，最后根据数组大小的奇偶性返回中间值。

class Solution {public:    double findMedianSortedArrays(vector<int>& nums1, vector<int>& nums2) {        bool isEven=((nums1.size()+nums2.size())%2==0);        if(nums1.size()==0&&nums2.size()==0)            return 0;            int Size=nums1.size()+nums2.size();            int size1=0,size2=0;            vector<int> nums;                while(size1<nums1.size()&&size2<nums2.size())                {                    if(nums1[size1]<nums2[size2])                    {                        nums.push_back(nums1[size1]);                        size1++;                    }                    else                     {                        nums.push_back(nums2[size2]);                        size2++;                    }                }                while(size2<nums2.size())                {                    nums.push_back(nums2[size2]);                    size2++;                }                while(size1<nums1.size())                {                    nums.push_back(nums1[size1]);                    size1++;                }            if(isEven)            {                double result=(double)(nums[Size/2-1]+nums[Size/2])/2;                return result;            }            else            {                double result=(double)nums[Size/2];                return result;            }    }};

29. Divide Two Integers

Divide two integers without using multiplication, division and mod operator.If it is overflow, return MAX_INT.

* 目标：在不调用乘法，除法，取余的情况下计算两个数相除的结果。

用“<<1”符号可代替乘法，申明变量temp等于除数，让temp每次乘2，看是否超过被除数，未超过则继续乘2，否则让被除数减去当前temp值，并重复上述步骤，同时记录商的值，直到余数小于除数。
注意：
1.重新声明除数和被除数为long long 型，避免溢出。
2.除数为零时，为溢出的情况，同时，当被除数为INT_MIN（-2147483648），除数为-1时也被判定为溢出，因为int类型无法表示+2147483648。
3.labs（）为长整型取绝对值函数

class Solution {public:    int divide(int dividend, int divisor) {        if(divisor==0||(dividend == INT_MIN && divisor == -1))            return INT_MAX;        if(dividend==0)            return 0;        int sign = 1;        if((dividend < 0) ^ (divisor < 0))                    sign = -1;        long long divid=labs(dividend);        long long divis=labs(divisor);        int result=0;            while (divid >= divis) {             long long temp = divis, mul = 1;            while (divid >= (temp << 1)) {                temp <<= 1;                mul<<= 1;            }            divid -= temp;            result += mul;        }        return sign == 1 ? result : -result;     }};

33. Search in Rotated Sorted Array

Suppose an array sorted in ascending order is rotated at some pivot unknown to you beforehand.(i.e., 0 1 2 4 5 6 7 might become 4 5 6 7 0 1 2).You are given a target value to search. If found in the array return its index, otherwise return -1.You may assume no duplicate exists in the array.Subscribe to see which companies asked this question.

目标：在数组中查找目标元素，但数组的排序可能是截断的。

最简单的方法是sort数组后在查找。更高效率的是用一个改版的二分查找。
1.找到数组中的最大值，该值将数组分为两半。
2.根据target的值决定在那一半使用二分查找。

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