算法设计与应用基础作业第二周

53. Maximum Subarray          

Descripyion：

Find the contiguous subarray within an array (containing at least one number) which has the largest sum.

For example, given the array [-2,1,-3,4,-1,2,1,-5,4],
the contiguous subarray [4,-1,2,1] has the largest sum = 6.

分析：这道题用分置的思想来做：

           和最大的子数组只能出现在三个位置：（从中间分开）左边、中间、（从中间分开）右边。

           所以分别求出（从中间分开）左边和最大的子数组、中间和最大的子数组、（从中间分开）右边和最大的子数组。

           然后比较三者中最大的子数组就是所求的子数组。

My C ++ code:

class Solution {public:    static int MaxSubSequence(vector <int>& A ,int N)    {    return MaxSubSum(A,0,N-1);    }    static int Max(int a, int b, int c)    {    return max(a, max(b,c));    }    static int MaxSubSum(vector <int>& A , int Left, int Right)    {    int MaxLeftSum,MaxRightSum;    int MaxLeftBorderSum,MaxRightBorderSum;    int LeftBorderSum,RightBorderSum;    int Center,i;    if(Left == Right)    {    return A[Left];    }    Center = (Left + Right)/2;    MaxLeftSum = MaxSubSum(A,Left,Center);    MaxRightSum = MaxSubSum(A,Center+1,Right);    MaxLeftBorderSum = A[Center];    LeftBorderSum = 0 ;    for(i = Center ;i >= Left;i--)    {    LeftBorderSum += A[i];    if(LeftBorderSum > MaxLeftBorderSum)    MaxLeftBorderSum = LeftBorderSum;    }    MaxRightBorderSum = A[Center + 1];    RightBorderSum = 0 ;    for(i = Center+1;i <= Right;i++)    {    RightBorderSum += A[i];    if(RightBorderSum > MaxRightBorderSum)    MaxRightBorderSum = RightBorderSum;    }        return Max(MaxLeftSum,MaxRightSum,MaxLeftBorderSum + MaxRightBorderSum);         }         int maxSubArray(vector<int>& nums) {        int n = nums.size() ;        return MaxSubSequence(nums , n) ;    }};

171. Excel Sheet Column Number          

Description:

Related to question Excel Sheet Column Title

Given a column title as appear in an Excel sheet, return its corresponding column number.

For example:

    A -> 1    B -> 2    C -> 3    ...    Z -> 26    AA -> 27    AB -> 28 

分析：这道题可以类比十进制数字的表示来做，例如：

     121 = 1*10^2+2*10^1+1*10^0 

     因此，就可以轻易地写出如下代码：

My C++ code:

class Solution {public:    int titleToNumber(string s) {        int total = 0 ;        int i = s.length() ;        for ( ; i > 0 ; i --)        {            total += (s[ i - 1 ] - 'A' + 1) * pow(26 , s.length() - i) ;        }        return total ;    }};