Day title series: Palindrome Partitioning II

Given A string s, partition s such that every substring of the partition are a palindrome.

Return the minimum cuts needed for a palindrome partitioning of s.

For example, given s = "aab" ,
Return 1 since the palindrome partitioning ["aa","b"] could be produced using 1 cut.

Description excerpt from ref http://blog.csdn.net/ljphhj/article/details/22573983

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Problem Solving Ideas:

We can turn this problem into a dynamically programmed DP problem.

First, we define a few variables, and do some of these quantities to explain! To facilitate understanding, these are pseudo-code!!!

Len = Str.length (); The length of the string

Int[] cuts = new Int[len + 1]; Cuts array, Cuts[i] indicates that a substring beginning with I to Len ends up to the minimum number of cuts required by test instructions (so that the final cuts[0] is the result we want) "Initialize cuts[i" = Len-i, because the worst case is that the substring that starts with I to Len ends up cutting the number Each character is cut once "

int[][] Matrix = new Int[len][len]; Sets an array of adjacency matrices, meaning that it represents: matrix[i][j] = True, which indicates that the sub-string sub (I, j) satisfies the palindrome string condition!

The criteria for determining whether matrix[i][j] satisfies a string of characters is:

Matrix[i+1][j-1] = = True (indicates sub (i+1,j-1) is satisfies palindrome string) && str[i] = = Str[j]

Or

J-i < 2 && str[i] = = Str[j] (that is, if j-i = = 1 o'clock, two characters are equal, if j-i = = 0 o'clock, the same character)

In both cases, we set matrix[i][j] to true to facilitate the next DP, and we can find the minimum number of cuts

Cuts[i] = Min{cuts[i], cuts[j+1] + 1}; State transition equation

So last cuts[0]-1 will be the smallest cut number of string STR!!!!

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Here the cut is initialized more specifically

 Public classSolution { Public intmincut (String s) {if(s==NULL|| S.length () <2)return0; intLen =s.length (); int[] Cut =New int[Len+1]; Boolean[] Matrix =New Boolean[Len][len];  for(inti=0; i<len;i++) {Cut[i]= Len-i;//Not len-i-1        }         for(inti=len-1;i>=0;i--){             for(intj=i;j<len;j++){                if((S.charat (i) ==s.charat (j)) && ((j-i<2) | | Matrix[i+1][j-1])) {Matrix[i][j]=true; Cut[i]= Math.min (cut[i],cut[j+1]+1); }            }        }        returnCut[0]-1; }}

Day title series: Palindrome Partitioning II