求字符串中不含重复字符的最长子串的长度

题目：求字符串最长不含重复字符的子串长度，如abcbec，就返回3.

分析：

利用动态规划(DP)原理，设字符串S的长度为n，考虑i...n-1这个后缀中符合条件的子串：首先需要记录两组数据，第一组数据是从i向右找到的最长不含重复字符的子串长度prefixlen[i]，第二组数据是在i...n-1后缀中符合条件的子串之起始和结束位置，分别用 maxlenstart[i]和maxlenend[i]表示，注意二组数据满足：maxlenstart[i]-maxlenend[i]+1>=prefixlen[i]，即从S[i]开始最左边的最长不含重复字符的子串长度与区间i...n-1中符合条件的子串长度可能相等；如果S[i]给prefixlen[i+1]增加一个字符长度，说明S[i]与右边长度为 prefixlen[i+1]个字符都不相同，这时需要分两种情况考虑，1）如果maxlenstart[i+1]==i+1即S[i+1...n-1] 中得到的最优子串的左边一个字符与S[i]也相邻，说明最大子串长度也应该增加1, 即设置maxlenstart[i]=i；注意此时如果prefixlen[i]总长度未在prefixlen[i+1]的基础上加1,说明第二组数据的长度不会比第一组数据长度大了，这是因为根据DP的计算过程，只要第二组数据与S[i]相邻，那么两组数据长度保证了前一步中是相等的；2）如果第二组数据不与S[i]相邻，则需要比较二组数据的长度了。

具体实现时需要注意边界条件，设置prefixlen
= 0; maxlenstart
= n; maxlenend
= n-1;。

代码：

/**
*
* @author ljs
* 2011-07-09
*
*/
public class LongestSubseqWithUniqueChars {

public static int solve(String str){
int n = str.length();
int[] prefixlen=new int[n+1];
prefixlen
= 0;
int[] maxlenstart = new int[n+1];
maxlenstart
= n;
int[] maxlenend = new int[n+1];
maxlenend
= n-1;

for(int i=n-1;i>=0;i--){
char c = str.charAt(i);
//caculate prefixlen[i] by prefixlen[i+1]
int j=0,k;
for(j=0,k=i+1;j<prefixlen[i+1];j++,k++){
if(c == str.charAt(k)){
break;
}
}
prefixlen[i] = j+1;

//caculate maxlenstart[i] and maxlenend[i]
maxlenstart[i] = maxlenstart[i+1];
maxlenend[i] = maxlenend[i+1];
if(maxlenstart[i+1] == i+1){
if(prefixlen[i] == prefixlen[i+1] + 1){
maxlenstart[i] = i;
}
}else{
//update the max len for i...n-1
if(maxlenend[i] - maxlenstart[i] + 1 < prefixlen[i]){
maxlenstart[i] = i;
maxlenend[i] = i + prefixlen[i] - 1;
}
}
}
return maxlenend[0] - maxlenstart[0] + 1;
}
public static void main(String[] args) {
String str = "abcbec";
int maxlen = LongestSubseqWithUniqueChars.solve(str);
System.out.format("max len of substring with unique chars for string \"%s\" = %d.%n",
str,maxlen);

str = "adabcbec";
maxlen = LongestSubseqWithUniqueChars.solve(str);
System.out.format("max len of substring with unique chars for string \"%s\" = %d.%n",
str,maxlen);

str = "abadadabbc";
maxlen = LongestSubseqWithUniqueChars.solve(str);
System.out.format("max len of substring with unique chars for string \"%s\" = %d.%n",
str,maxlen);

str = "ffdeefghff";
maxlen = LongestSubseqWithUniqueChars.solve(str);
System.out.format("max len of substring with unique chars for string \"%s\" = %d.%n",
str,maxlen);

str = "abcbecghijkl";
maxlen = LongestSubseqWithUniqueChars.solve(str);
System.out.format("max len of substring with unique chars for string \"%s\" = %d.%n",
str,maxlen);

}

}

测试输出：

max len of substring with unique chars for string "abcbec" = 3.
max len of substring with unique chars for string "adabcbec" = 4.
max len of substring with unique chars for string "abadadabbc" = 3.
max len of substring with unique chars for string "ffdeefghff" = 4.
max len of substring with unique chars for string "abcbecghijkl" = 9.

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