面试笔试杂项积累-leetcode 61-65

61.61-Rotate List-Difficulty:Medium

Given a list, rotate the list to the right by k places, where k is non-negative.

For example:
Given 1->2->3->4->5->NULL and k = 2,
return 4->5->1->2->3->NULL.

思路

给一个链表返回旋转k个之后的链表。很简单，先遍历一遍使head接在end上，顺便求出长度，然后直接在length-k处断掉即可

/** * Definition for singly-linked list. * public class ListNode { *     public int val; *     public ListNode next; *     public ListNode(int x) { val = x; } * } */public class Solution {    public ListNode RotateRight(ListNode head, int k) {        if(head==null||k == 0)        return head;                ListNode last = head;        ListNode temp;        int i = 1;        while (last.next != null)        {            last = last.next;            ++i;        }        while(i <=k)     k = k-i;        i = i - k;        last.next = head;        last = head;        while (i > 1)        {            last = last.next;            --i;        }        temp = last.next;        last.next = null;        return temp;    }}

62.62-Unique Paths-Difficulty:Medium

A robot is located at the top-left corner of a m x n grid (marked 'Start' in the diagram below).

The robot can only move either down or right at any point in time. The robot is trying to reach the bottom-right corner of the grid (marked 'Finish' in the diagram below).

How many possible unique paths are there?

Above is a 3 x 7 grid. How many possible unique paths are there?

Note: m and n will be at most 100.

思路

找到左上角到右下角所有可能的路径数量。

使用动态规划能轻松解决

对于格点(i,j)。由于只能从上格点(i-1,j)或左格点(i,j-1)到达，并且两者路径是不重复的

因此path[i][j] = path[i-1][j]+path[i][j-1]

附上动态规划教程链接：

五大常用算法之二：动态规划算法

public class Solution {      public int UniquePaths(int m, int n)    {              int[,] fin = new int[m, n];        for (int i = 0; i < m; i++)            for (int j = 0; j < n; j++)                fin[i, j] = 1;        for (int i = 1; i < m; i++)        {            for (int j = 1; j < n; j++)            {                fin[i, j] = fin[i - 1, j] + fin[i, j - 1];            }        }        return fin[m - 1, n - 1];    }}

63.63-Unique Paths II-Difficulty:Medium

Follow up for "Unique Paths":

Now consider if some obstacles are added to the grids. How many unique paths would there be?

An obstacle and empty space is marked as 1 and0 respectively in the grid.

For example,

There is one obstacle in the middle of a 3x3 grid as illustrated below.

[  [0,0,0],  [0,1,0],  [0,0,0]]

The total number of unique paths is 2.

Note: m and n will be at most 100.

思路

和上一题大体相同，但加入了障碍物的情况，有障碍物的地方不能到达，如果终点本身为障碍物则不能到达

public class Solution {    public int UniquePathsWithObstacles(int[,] obstacleGrid) {      int m = obstacleGrid.GetLength(0);        int n = obstacleGrid.GetLength(1);        if(obstacleGrid[m - 1, n - 1] == 1)        return 0;        int[,] fin = new int[m, n];        for (int i = 0; i < m; i++)            for (int j = 0; j < n; j++)                fin[i, j] = 0;        for (int i = 0; i < m; i++)        {            if (obstacleGrid[i,0] != 1)                fin[i,0] = 1;            else                break;        }        for (int i = 0; i < n; i++)        {            if (obstacleGrid[0,i] != 1)                fin[0,i] = 1;            else                break;        }        for (int i = 1; i < m; i++)        {            for (int j = 1; j < n; j++)            {                if (obstacleGrid[i,j] == 1)                    fin[i,j] = 0;                else                fin[i, j] = fin[i - 1, j] + fin[i, j - 1];            }        }        return fin[m - 1, n - 1];    }}

64.64-Minimum Path Sum-Difficulty:Medium

Given a m x n grid filled with non-negative numbers, find a path from top left to bottom right whichminimizes the sum of all numbers along its path.

Note: You can only move either down or right at any point in time.

思路

和上两道题相近，返回左上到右下最小的路径节点和，依旧动态规划

public class Solution {    public int MinPathSum(int[,] grid) {        int m = grid.GetLength(0);        int n = grid.GetLength(1);        if(m == 1&&n ==1)        return grid[0,0];        int[,] fin = new int[m, n];        Array.Copy(grid, fin, grid.Length);        for (int i = 2; i < n; i++)        {            fin[0, i] += fin[0, i - 1];        }        for (int i = 2; i < m; i++)        {            fin[i, 0] += fin[i - 1, 0];        }        for (int i = 1; i < m; i++)        {            for (int j = 1; j < n; j++)            {                fin[i, j] += fin[i - 1, j] > fin[i, j - 1] ? fin[i, j - 1] : fin[i - 1, j];            }        }        return fin[m - 1, n - 1]+fin[0,0];    }}

65.65-Minimum Path Sum-Difficulty:Hard

Validate if a given string is numeric.

Some examples:
"0" => true
" 0.1 " => true
"abc" => false
"1 a" => false
"2e10" => true

Note: It is intended for the problem statement to be ambiguous. You should gather all requirements up front before implementing one.

思路

 判断字符串是不是数字，不用考虑前后缀空格。

需要注意：

1.小数的"."的位置

2."e"存在的位置和次数

3.正负号

public class Solution {    public bool IsNumber(string s) {          s = s.TrimEnd(' ');        s = s.TrimStart(' ');        if (s.Length == 0)            return false;        int start = 0;        if (s[0] == 43 || s[0] == 45)            start = 1;        bool isDecimals = false;        bool isE = false;        for (int i = start; i < s.Length; i++)        {            if (!isDecimals && !isE)            {                if (s[i] == 46)                {                    if (i == s.Length - 1 && i <= start)                    { return false; }                    isDecimals = true;                    continue;                }            }            if (s[i] < 48 || s[i] > 57)            {                if (!isE )                if (s[i] == 101 && i > start && i < s.Length-1)                {                    if(s[i-1] == 46&&i-1 == start)                        return false;                    isE = true;            if ((s[i+1] == 43 || s[i+1] == 45)&&i+1<s.Length-1)            ++i;                                        continue;                }                return false;            }        }        return true;    }}