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I have applied the lm() to a data set. The independent variables are categorical. First, I use lm() with intercept and I got the next results:

> model <- lm(y ~ factor(x))
> summary(model)

Call:
lm(formula = y ~ factor(x))

Residuals:
    Min      1Q  Median      3Q     Max 
-5.3085 -1.8132 -0.4136  1.4323 11.2480 

Coefficients:
              Estimate Std. Error t value Pr(>|t|)    
(Intercept)     9.3085     0.4064  22.907   <2e-16 ***
factor(x)0.75   0.1435     0.6896   0.208    0.836    
factor(x)1.5    0.9062     0.6272   1.445    0.151    
factor(x)3      0.9040     0.6989   1.293    0.198    
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 

Residual standard error: 2.786 on 126 degrees of freedom
Multiple R-squared: 0.0238, Adjusted R-squared: 0.0005601 
F-statistic: 1.024 on 3 and 126 DF,  p-value: 0.3844

In the second model I don't use intercept:

> model.1 <- lm(y ~ factor(x) - 1)
> summary(model.1)

Call:
lm(formula = y ~ factor(x) - 1)

Residuals:
    Min      1Q  Median      3Q     Max 
-5.3085 -1.8132 -0.4136  1.4323 11.2480 

Coefficients:
              Estimate Std. Error t value Pr(>|t|)    
factor(x)0.25   9.3085     0.4064   22.91   <2e-16 ***
factor(x)0.75   9.4520     0.5572   16.96   <2e-16 ***
factor(x)1.5   10.2147     0.4778   21.38   <2e-16 ***
factor(x)3     10.2125     0.5687   17.96   <2e-16 ***
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 

Residual standard error: 2.786 on 126 degrees of freedom
Multiple R-squared: 0.9267, Adjusted R-squared: 0.9243 
F-statistic:   398 on 4 and 126 DF,  p-value: < 2.2e-16

If I don't understand the difference between the R-squared value of them? Could I accept the second one as a fit model?

Would somebody help me to understand this problem?

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It's a well-known problem. However, you could use

R_squared <- (corr(y, fitted(model.1)))^2

to get a comparable $R^2$. In fact, $\text{corr}^2(y,\hat{y})$ is equivalent to the R squared printed by summary when there is an intercept (see M. Verbeek, §2.4).

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