41. First Missing Positive *HARD*

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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.

Your algorithm should run in O(n) time and uses constant space.

int firstMissingPositive(vector<int>& nums) {
    int n = nums.size(), t, i;
    for(i = 0; i < n; i++)
    {
        t = nums[i];
        while(t > 0 && t < n && nums[t-1] != t)
        {
            swap(nums[i], nums[t-1]);
            t = nums[i];
        }
    }
    for(i = 0; i < n; i++)
    {
        if(nums[i] != i+1)
            return i+1;
    }
    return n+1;
}

Idea:
*
* We can move the num to the place whcih the index is the num.
*
* for example, (considering the array is zero-based.
* 1 => A[0], 2 => A[1], 3=>A[2]
*
* Then, we can go through the array check the i+1 == A[i], if not ,just return i+1;

41. First Missing Positive *HARD*

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原文地址：http://www.cnblogs.com/argenbarbie/p/5245569.html