Longest Consecutive Sequence 数组连续数字的情况

Longest Consecutive Sequence

Given an unsorted array of integers, find the length of the longest consecutive elements sequence.

For example,

Given

[100, 4, 200, 1, 3, 2]

,

The longest consecutive elements sequence is

[1, 2, 3, 4]

. Return its length:

4

.

寻找最大的连续数组的长度，利用两个set数组，一个数组存储已经被访问过的数组，如果已经被访问过了，那么就不需要再被访问了

class Solution {
public:
    int longestConsecutive(vector<int>& nums) {
        if(nums.empty())
        {
            return 0;
        }
        unordered_set<int> existSet;
        unordered_set<int> visitedSet;
        int maxLength = 0;
        for(int i = 0; i < nums.size(); i++)
            existSet.insert(nums[i]);
        for(int i = 0; i < nums.size(); i++)
        {
            int length = 0;
            if(visitedSet.count(nums[i]))
            {
                continue;
            }
            else
            {
                visitedSet.insert(nums[i]);
                length++;
                int left = nums[i];
                int right = nums[i];
                while(existSet.count(--left))
                {
                    visitedSet.insert(left);
                    length++;
                }
                while(existSet.count(++right))//<必须使用�前＋＋
                {
                    visitedSet.insert(right);
                    length++;
                }
                maxLength = max(maxLength,length);
            }
        }
        return maxLength;
    }
};

First Missing Positive

Given an unsorted integer array, find the first missing positive integer.

For example,

Given

[1,2,0]

return

3

,

and

[3,4,-1,1]

return

2

.

连续数组求最小的miss的数据，还是利用前面介绍的方法，还是利用两个不同的set，其中一个表示已经访问过了的情况，只观察比当前大的数据，

如果已经访问过了就不访问了

class Solution {
public:
    int firstMissingPositive(vector<int>& nums) {
        if(nums.empty())
        {
            return 1;
        }
        int missVal = INT_MAX;
        int minVal = 1;
        unordered_set<int> numSet;
        unordered_set<int> visitedSet;
        for_each(nums.begin(),nums.end(),[&numSet](int x)
        {
            if(x >= 0)
                numSet.insert(x);
        });
        for(int i = 0; i < nums.size(); i++)
        {
            if(nums[i] < 0 || visitedSet.count(nums[i]))
            {
                continue;
            }
            int right = nums[i];
            while(numSet.count(++right))
            {
                visitedSet.insert(right);
            }
            missVal = min(missVal,right);
        }
        if(numSet.count(1))
            return missVal;
        else
        {
            return 1;
        }
    }
};