128. Longest Consecutive Sequence

方法一：并查集

class UF {
public:
    UF(int n) {
        for (int i = 0; i < n; ++i)
            roots.push_back(i);
    }
    
    void Union(int p, int q) {
        int pRoot = findRoot(p);
        int qRoot = findRoot(q);
        if (qRoot == pRoot)
            return;
        else {
            roots[pRoot] = qRoot;
        }
    }
    
    int findRoot(int p) {
        while (roots[p] != p)
            p = roots[p];
        return p;
    }
    
    bool isUnion(int p, int q) {
        return findRoot(p) == findRoot(q);
    }
private:
    vector<int> roots;
};

class Solution {
public:
    int longestConsecutive(vector<int>& nums) {
        UF uf(nums.size());
        unordered_map<int, int> m;
        unordered_map<int, int> rootSize;
        for (int i = 0; i < nums.size(); ++i) {//构造过程中，每个index的root可能会变
            if (m.find(nums[i]) != m.end())
                continue;
            m[nums[i]] = i;
            if (m.find(nums[i] - 1) != m.end())
                uf.Union(i, m[nums[i]-1]);
            if (m.find(nums[i] + 1) != m.end())
                uf.Union(i, m[nums[i]+1]);
        }
        for (int i = 0; i < nums.size(); ++i)
            rootSize[uf.findRoot(i)]++;
        int maxVal = 0;
        for (auto& item : rootSize)
            maxVal = max(maxVal, item.second);
        return maxVal;
    }
};

方法二：
class Solution {
public:
int longestConsecutive(vector& nums) {
unordered_map<int, int> m;
int ret = 0;
for (int num : nums) {
if (m.find(num) != m.end())
continue;
int leftLen = (m.find(num-1) != m.end() ? m[num-1] : 0);
int rightLen = (m.find(num+1) != m.end() ? m[num+1] : 0);
int len = 1 + leftLen + rightLen;
ret = max(len, ret);
m[num] = len;//为什么这行不能去掉？哦哦因为要把这个num加入到map里面，表示遇过了。但这个值没有意义。写个0也行
m[num-leftLen] = len;
m[num+rightLen] = len;
}
return ret;
}
};
//一段连续的数字的两端的数字在map中对应着这段的长度。