shop. Inspired by the color-coded bands on resistors, he is painting
a series of long, narrow canvases with bold colored stripes. However,
Karel is lazy and wants to minimize the number of brush strokes it takes
to paint each canvas.
Abbreviating each color to a single uppercase letter, Karel would
write the stripe pattern red-green-blue-green-red as "RGBGR"
(quotes added for clarity). It would take him three brush strokes to
paint this pattern.
The first stroke would cover the entire canvas in red (RRRRR).
The second stroke would leave a band of red on either side and fill
in the rest with green (RGGGR).
The final brush stroke would fill in the blue stripe in the center
(RGBGR).
Given a stripe pattern stripes as a String, calculate the minimum number of
brush strokes required to paint that pattern.
DEFINITION
Class:StripePainter
Method:minStrokes
Parameters:String
Returns:int
Method signature:int minStrokes(String stripes)
NOTES
-The blank canvas is an ugly color and must not show through.
CONSTRAINTS
-stripes will contain only uppercase letters ('A'-'Z', inclusive).
-stripes will contain between 1 and 50 characters, inclusive.
EXAMPLES
0)
"RGBGR"
Returns: 3
Example from introduction.
1)
"RGRG"
Returns: 3
This example cannot be done in two strokes, even though there are only two colors.
Suppose you tried to paint both red stripes in one stroke, followed by both green stripes
in one stroke. Then the green stroke would cover up the second red stripe. If you tried
to paint both green stripes first, followed the red stripes, then the red stroke would
cover up the first green stripe.
2)
"ABACADA"
Returns: 4
One long stroke in color 'A', followed by one stroke each in colors 'B', 'C', and 'D'.
3)
"AABBCCDD CCBB AABBCCDD"
Returns: 7
4)
"BECBBDDEEBABDCADEAAEABCACBDBEECDEDEACACCBEDABEDADD"
Returns: 26
class StripePainter 
{
static int[][] dp;//dp[i][j] 琛ㄧず浠巌-j鐨勮寖鍥村唴 鏈灝忕殑娑傝壊嬈℃暟 static int n;
public int minStrokes(String s) { int res = 0; dp = new int[s.length()][s.length()]; n = s.length(); for (int i = 0; i < n; i++) { dp[i][i] = 1;
} for (int i = 1; i < n; i++) { for (int j = 0, t = j + i; j < n && t < n; j++, t = j + i) { int min = Integer.MAX_VALUE; for (int k = j; k < t; k++) { int tmp = dp[j][k] + dp[k + 1][t]; if (s.charAt(j) == s.charAt(t)) tmp -= 1; if (min > tmp) min = tmp; } dp[j][t] = min; } } return dp[0][s.length() - 1]; }
}]]>