求最长公共子序列-DP问题

Longest common subsequence problem

The longest common subsequence (LCS) problem is the problem of finding the longest subsequence common to all sequences in a set of sequences (often just two sequences). It differs from the longest common substring problem: unlike substrings, subsequences are not required to occupy consecutive positions within the original sequences.

Example

  • LCS for input Sequences ABCDGH and AEDFHR is ADH of length 3.
  • LCS for input Sequences AGGTAB and GXTXAYB is GTAB of length 4

以s1={1,3,4,5,6,7,7,8},s2={3,5,7,4,8,6,7,8,2}为例。

1.构建二维矩阵A[n+1][n+1] s1为列,s2为行

(1)A[0][any]=0,A[any][0]=0

(2)if(s1[columnIndex-1]==s2[rowIndex-1])则A[rowIndex][columnIndex]=A[rowIndex-1][columnIndex-1]+1;

         else  A[rowIndex][columnIndex]=max(A[rowIndex][columnIndex-1],A[rowIndex-1][columnIndex])

 (3)由A[8][9]可知最大子序列长度。

图如下:

 

 2.针对构建的二维矩阵求最长子序列

(1)当columnIndex>0或者rowIndex>0时。

(2)如果s1[columnIndex-1]==s2[rowIndex-1]则longestSequence.unshift(s1[columnIndex-1]).rowIndex--;columnIndex--.

(3)如果不等,

              则1.如果A[rowIndex][columnIndex]==A[rowIndex][columnIndex-1];columnIndex--;//向左

      2.否则 rowIndex--;

 

 

代码如下:

 

/**
 * @param {string[]} set1
 * @param {string[]} set2
 * @return {string[]}
 */
export default function longestCommonSubsequence(set1, set2) {
  // Init LCS matrix.
  const lcsMatrix = Array(set2.length + 1).fill(null).map(() => Array(set1.length + 1).fill(null));

  // Fill first row with zeros.
  for (let columnIndex = 0; columnIndex <= set1.length; columnIndex += 1) {
    lcsMatrix[0][columnIndex] = 0;
  }

  // Fill first column with zeros.
  for (let rowIndex = 0; rowIndex <= set2.length; rowIndex += 1) {
    lcsMatrix[rowIndex][0] = 0;
  }

  // Fill rest of the column that correspond to each of two strings.
  for (let rowIndex = 1; rowIndex <= set2.length; rowIndex += 1) {
    for (let columnIndex = 1; columnIndex <= set1.length; columnIndex += 1) {
      if (set1[columnIndex - 1] === set2[rowIndex - 1]) {
        lcsMatrix[rowIndex][columnIndex] = lcsMatrix[rowIndex - 1][columnIndex - 1] + 1;
      } else {
        lcsMatrix[rowIndex][columnIndex] = Math.max(
          lcsMatrix[rowIndex - 1][columnIndex],
          lcsMatrix[rowIndex][columnIndex - 1],
        );
      }
    }
  }

  // Calculate LCS based on LCS matrix.
  if (!lcsMatrix[set2.length][set1.length]) {
    // If the length of largest common string is zero then return empty string.
    return [''];
  }

  const longestSequence = [];
  let columnIndex = set1.length;
  let rowIndex = set2.length;

  while (columnIndex > 0 || rowIndex > 0) {
    if (set1[columnIndex - 1] === set2[rowIndex - 1]) {
      // Move by diagonal left-top.
      longestSequence.unshift(set1[columnIndex - 1]);
      columnIndex -= 1;
      rowIndex -= 1;
    } else if (lcsMatrix[rowIndex][columnIndex] === lcsMatrix[rowIndex][columnIndex - 1]) {
      // Move left.
      columnIndex -= 1;
    } else {
      // Move up.
      rowIndex -= 1;
    }
  }

  return longestSequence;
}

 


posted @ 2019-03-18 11:31  Archer-Fang  阅读(226)  评论(0编辑  收藏  举报