[LeetCode]Add Binary&Climbing Stairs&Remove Duplicates from Sorted List

今天写得蛮快的，来个三连发。

Add Binary

Given two binary strings, return their sum (also a binary string).

For example,
a = "11"
b = "1"
Return "100".

• 题意：二进制加法.
• 解法：把两个串都倒过来(低位先加).然后模拟一下就好了.注意最后也要进位就好.

class Solution {    static int getIntByIndxInString(const std::string& rkStr, int iIndx) {        return iIndx < (int)rkStr.size() ? rkStr[iIndx] - '0' : 0;    }public:    string addBinary(string a, string b) {        reverse(a.begin(), a.end());        reverse(b.begin(), b.end());        string kResult;        int iAddValue = 0;        for (int i = 0; i < (int)a.size() || i < (int)b.size(); ++i) {            int iAddSum = getIntByIndxInString(a, i) + getIntByIndxInString(b, i) + iAddValue;//这里是整数相加结果只有0 1 2            kResult.push_back((iAddSum & 1) + '0');//iAddSum&1结果为0和1，再加上'0'就是字符串的0和1了。            iAddValue = iAddSum > 1;//如果和为2说明有进位.        }        if (iAddValue)            kResult.push_back('1');        reverse(kResult.begin(), kResult.end());        return kResult;    }};

Climbing Stairs

You are climbing a stair case. It takes n steps to reach to the top.

Each time you can either climb 1 or 2 steps. In how many distinct ways can you climb to the top?

• 题意：走梯子，一个普通的递推.
• 对于第i个梯子有前一步和前两步都能到达它，所以方案为前两个的和，方程式为dp[i]=dp[i-1]+dp[i-2](i>1).

//用a,b表示c的前两步class Solution {public:    int climbStairs(int n) {        int a = 0, b = 1, c = 1;        for (; n--;) {            c = a + b;            a = b;            b = c;        }        return c;    }};

Remove Duplicates from Sorted List

Given a sorted linked list, delete all duplicates such that each element appear only once.

For example,
Given 1->1->2, return 1->2.
Given 1->1->2->3->3, return 1->2->3.

• 题意：有序链表的去重
• 解法：跟这里差不多，不过本题不能直接访问上一个节点，所以得把上一个节点也存下来在循环的时候.

/** * Definition for singly-linked list. * struct ListNode { *     int val; *     ListNode *next; *     ListNode(int x) : val(x), next(NULL) {} * }; */class Solution {public:    ListNode* deleteDuplicates(ListNode* head) {        if (!head) return nullptr;        ListNode* pkResult = head;        ListNode* pkMove = pkResult;        ListNode* pkLastNode = head;        for (head = head->next; head; head = head->next, pkLastNode = pkLastNode->next) {            if (head->val != pkLastNode->val) {                pkMove->next = head;                pkMove = pkMove->next;            }        }        pkMove->next = nullptr;        return pkResult;    }};

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