Title: Given a sequence of length n, m is a number of labels (label value is expressed as 1,2, ...., m), emission probability matrix E (n * m), where E [i] [j] denotes the i-th emission probability is predicted word label j, the transition probability matrix T (m * m), where T [i] [j] is transferred to the label i j labels the transition probability. He asked to return to the optimal sequence annotation result (an array res, res [i] denotes the i-th word label value marked).
public class Solution { public int[] getBestPath(int m, int n, double[][] E, double[][] T) { double[][] dp = new double[n + 1][m + 1]; int[][] his = new int[n + 1][m + 1]; for (int j = 1; j <= m; j++) { dp[1][j] = E[1][j]; } for (int i = 2; i <= n; i++) { for (int j = 1; j <= m ; j++) { for (int k = 1; k <= m; k++) { double score = dp[i - 1][k] + T[k][j] + E[i][j]; if (score > dp[i][j]) { dp[i][j] = score; his[i][j] = k; } } } } int label = 0; double min = Double.MIN_VALUE; for (int j = 1; j <= m; j++) { if (dp[n][j] > min) { label = j; min = dp[n][j]; } } int[] res = new int[n + 1]; for (int i = n; i >= 1; i--) { res[i] = label; label = his[i][label]; } return res; } }