Leetcode日记（11）

Combination Sum II

问题描述

    Given a collection of candidate numbers (C) and a target number (T), find all unique combinations in C where the candidate numbers sums to T.
Each number in C may only be used once in the combination.
    Note:
    • All numbers (including target) will be positive integers.
    • The solution set must not contain duplicate combinations.
    For example, given candidate set [10, 1, 2, 7, 6, 1, 5] and target 8,
    A solution set is:

[  [1, 7],  [1, 2, 5],  [2, 6],  [1, 1, 6]]

分析

    采用回溯算法，注意每个值只能使用一次，并且所有记过都是唯一的。

解答

class Solution {public:    vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {        sort(candidates.begin(), candidates.end());        vector<vector<int>> result;        vector<int> combination;        combinationsum(result, combination, candidates, target, 0);        return result;    }    void combinationsum(vector<vector<int>> &res, vector<int> &com, vector<int> &arr, int tar, int begin)    {        if (!tar)        {            res.push_back(com);            return ;        }        for (int i = begin; i < arr.size() && tar >= arr[i]; i++)        {            com.push_back(arr[i]);            combinationsum(res, com, arr, tar - arr[i], i+1);            com.pop_back();            while (i + 1 < arr.size() && arr[i] == arr[i + 1])                i++;        }    }};

Multiply Strings

问题描述

     Given two non-negative integers num1 and num2 represented as strings, return the product of num1 and num2.
    Note:

1. The length of both num1 and num2 is < 110.
2. Both num1 and num2 contains only digits 0-9.
3. Both num1 and num2 does not contain any leading zero.
4. You must not use any built-in BigInteger library or convert the inputs to integer directly.

分析

    由于字符串的长度未知，且可能超出长整型的长度，所以我们运用手动乘法的方式来解决。

解答

class Solution {public:    string multiply(string num1, string num2) {        string sum(num1.size() + num2.size(), '0');        for (int i = num1.size() - 1; 0 <= i; i--)         {            int carry = 0;            for (int j = num2.size() - 1; 0 <= j; j--)             {                int tmp = (sum[i + j + 1] - '0') + (num1[i] - '0') * (num2[j] - '0') + carry;                sum[i + j + 1] = tmp % 10 + '0';                carry = tmp / 10;            }            sum[i] += carry;        }        size_t startpos = sum.find_first_not_of("0");        if (string::npos != startpos)         {            return sum.substr(startpos);        }        return "0";    }};

Permutations

问题描述

     Given a collection of distinct numbers, return all possible permutations.
    For example,
    [1,2,3] have the following permutations:

[  [1,2,3],  [1,3,2],  [2,1,3],  [2,3,1],  [3,1,2],  [3,2,1]]

分析

    使用回溯算法设计，比较经典的题目。

解答

class Solution {public:    vector<vector<int>> permute(vector<int>& nums) {        vector<vector<int>> result;        getres(result, nums, 0);        return result;    }    void getres(vector<vector<int>> &res, vector<int> &nums, int start)    {        if (start >= nums.size())        {            res.push_back(nums);            return ;        }        for (int i = start; i < nums.size(); i++)        {            swap(nums[start], nums[i]);            getres(res, nums, start + 1);            swap(nums[start], nums[i]);        }    }};

Permutations II

问题描述

     Given a collection of numbers that might contain duplicates, return all possible unique permutations.
    For example,
    [1,1,2] have the following unique permutations:

[  [1,1,2],  [1,2,1],  [2,1,1]]

分析

    和上一问题相同，需要注意数字出现重复的情况。

解答

class Solution {public:    vector<vector<int> > permuteUnique(vector<int> &num) {        sort(num.begin(), num.end());        vector<vector<int> >res;        getres(res, num, 0, num.size());        return res;    }    void getres(vector<vector<int>> &res, vector<int> num, int start, int end)     {        if (start == end - 1) {            res.push_back(num);            return;        }        for (int k = start; k < end; k++)         {            if (start != k && num[start] == num[k])             continue;            swap(num[start], num[k]);            getres(res, num, start + 1, end);        }    }};