The function $f(n)$ satisfies $f(0)=0$, $f(n)=n-f \left( f(n-1) \right)$, $n=1,2,3 \cdots$. Find all polynomials $g(x)$ with real coefficient such that
\[ f(n)= [ g(n) ], \qquad n=0,1,2 \cdots \]
Where $[ g(n) ]$ denote the greatest integer that does not exceed $g(n)$.
1. **Identify the positive root of the quadratic equation:**
   \[
   x^2 - x - 1 = 0
   \]
   The roots of this equation are given by the quadratic formula:
   \[
   x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
   \]
   Substituting \(a = 1\), \(b = -1\), and \(c = -1\):
   \[
   x = \frac{1 \pm \sqrt{1 + 4}}{2} = \frac{1 \pm \sqrt{5}}{2}
   \]
   The positive root is:
   \[
   \alpha = \frac{1 + \sqrt{5}}{2}
   \]

2. **Verify the properties of \(\alpha\):**
   \[
   \alpha = 1 + \frac{1}{\alpha}
   \]
   and
   \[
   \frac{1}{\alpha^2} + \frac{1}{\alpha} = 1
   \]
   These properties can be derived from the quadratic equation and its roots.

3. **Prove by induction that \(f(n) = \left\lfloor \frac{n+1}{\alpha} \right\rfloor\):**
   - **Base case:** For \(n = 0\),
     \[
     f(0) = 0 = \left\lfloor \frac{0+1}{\alpha} \right\rfloor = \left\lfloor \frac{1}{\alpha} \right\rfloor = 0
     \]
   - **Inductive step:** Assume \(f(k) = \left\lfloor \frac{k+1}{\alpha} \right\rfloor\) for all \(k \leq n-1\). We need to show that:
     \[
     f(n) = \left\lfloor \frac{n+1}{\alpha} \right\rfloor
     \]
     Given \(f(n) = n - f(f(n-1))\), let \(k = \left\lfloor \frac{n}{\alpha} \right\rfloor\). Then:
     \[
     f(n-1) = \left\lfloor \frac{n}{\alpha} \right\rfloor = k
     \]
     We need to prove:
     \[
     f(n) = n - f(k)
     \]
     Since \(f(k) = \left\lfloor \frac{k+1}{\alpha} \right\rfloor\), we have:
     \[
     f(n) = n - \left\lfloor \frac{k+1}{\alpha} \right\rfloor
     \]

4. **Consider two cases for the value of \(k\):**
   - **Case 1:** \(k < \frac{n}{\alpha} < \frac{n+1}{\alpha} < k+1\)
     \[
     \left\lfloor \frac{n+1}{\alpha} \right\rfloor = k
     \]
     We need to show:
     \[
     k + \left\lfloor \frac{k+1}{\alpha} \right\rfloor = n
     \]
     Since \(\left\lfloor \frac{k+1}{\alpha} \right\rfloor\) is an integer, we have:
     \[
     k + \left\lfloor \frac{k+1}{\alpha} \right\rfloor > n-1 \quad \text{and} \quad k + \left\lfloor \frac{k+1}{\alpha} \right\rfloor < n+1
     \]
     This implies:
     \[
     k + \frac{k+1}{\alpha} - 1 > n-1 \quad \text{and} \quad k + \frac{k+1}{\alpha} < n+1
     \]
     Simplifying, we get:
     \[
     k\alpha + \frac{1}{\alpha} - 1 > n-1 \quad \text{and} \quad k\alpha + \frac{1}{\alpha} < n+1
     \]
     Both inequalities hold true by our assumption.

   - **Case 2:** \(k < \frac{n}{\alpha} < k+1 < \frac{n+1}{\alpha}\)
     \[
     \left\lfloor \frac{n+1}{\alpha} \right\rfloor = k+1
     \]
     We need to show:
     \[
     k + 1 + \left\lfloor \frac{k+1}{\alpha} \right\rfloor = n
     \]
     Since \(\left\lfloor \frac{k+1}{\alpha} \right\rfloor\) is an integer, we have:
     \[
     k + 1 + \left\lfloor \frac{k+1}{\alpha} \right\rfloor > n-1 \quad \text{and} \quad k + 1 + \left\lfloor \frac{k+1}{\alpha} \right\rfloor < n+1
     \]
     This implies:
     \[
     (k+1)\alpha - 1 > n-1 \quad \text{and} \quad (k+1)\alpha < n+1
     \]
     Both inequalities hold true by our assumption.

5. **Determine the polynomial \(g(x)\):**
   Since \(f(n) = \left\lfloor \frac{n+1}{\alpha} \right\rfloor\), we assume \(g(x) = ux + v\). We need:
   \[
   \left| n \left( \frac{1}{\alpha} - u \right) + \frac{1}{\alpha} - v \right| < 1
   \]
   Clearly, \(u = \frac{1}{\alpha}\). Using the density of \(\left\{ \frac{n}{\alpha} \right\}\) in \((0,1)\), we find \(v = \frac{1}{\alpha}\). Thus:
   \[
   g(n) = \frac{n+1}{\alpha}
   \]

The final answer is \(\boxed{\frac{n+1}{\alpha}}\)