[LeetCode]217-Contains Duplicate 几种解法

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数组重复元素查找：

解法一：sort 后相邻位比较 40ms

class Solution {public:    bool containsDuplicate(vector<int>& nums) {                if(nums.size() <= 1) return false;        sort(nums.begin(),nums.end());        for(int i = 0; i<nums.size()-1;i++){            if(nums[i] == nums[i+1]) return true;        }        return false;            }};

解法二：采用STL的 unordered_set 基本hash思想 52ms

class Solution {public:    bool containsDuplicate(vector<int>& nums) {        unordered_set<int> st;        for (auto n : nums) {            if (st.find(n) != st.end()) return true;            else st.insert(n);        }        return false;    }};

解法三：map记录已存在的int

class Solution {public:    bool containsDuplicate(vector<int>& nums) {     //runtime:104ms        if(nums.size()<=1)return false;        map<int,bool> m;        for(int i=0;i<nums.size();i++){            if(m[nums[i]]==true)return true;            m[nums[i]]=true;        }        return false;    }};

解法四：自己手写一个快排算法 36ms

class Solution {public:    bool containsDuplicate(vector<int>& nums) {        int len = nums.size();        if (len < 1) {            return false;        }        quick_sort(nums, 0, len);        int prev = nums[0];        for (int i=1; i<len; i++) {            if (nums[i] == prev) {                return true;            }            prev= nums[i];        }        return false;    }        void quick_sort(vector<int>& nums, int start, int end) {        if (start >= end) {            return;        }        int di = partition(nums, start, end);        quick_sort(nums, start, di);        quick_sort(nums, di + 1, end);    }        int partition(vector<int>& nums, int start, int end) {        if (start >= end) {            return -1;        }        end--;        swap(nums[(start + end) / 2], nums[end]);        int pv = nums[end];        int di = start;        for (int i=start; i<end; i++) {            if (nums[i] < pv) {                swap(nums[i], nums[di++]);            }        }        swap(nums[end], nums[di]);        return di;    }};

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