O(n) solution in Java, 0ms, statistic mathod

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    public int uniquePaths(int m, int n) {
        double result = 1;
            int temp = n;
            n = m;
            m = temp;
        for(int i=0;i<m-1;i++){
            result = (result*(m+n-2-i))/(i+1);
        return (int)result;

    Totally m+n-2 steps. We have to choose m-1 steps as down (or n-1 steps as right, we should choose the smaller one), regardless of the sequence of these m-1 steps..

    Based on statistic knowledge, it will be (m+n-2)!/((m-1)!*(n-1)!).

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