LSTM+CRF维特比解码过程

题目：给定长度为n的序列，标签数为m（标签值表示为1,2,....,m），发射概率矩阵E（n * m），其中E[i][j]表示第i个词预测为标签j的发射概率，转移概率矩阵T(m*m)，其中T[i][j]为标签i转移到标签j的转移概率。要求返回最优的序列标注结果（一个数组res，res[i]表示第i个词标注的标签值）。

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;
    }
}

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来源：oschina

链接：https://my.oschina.net/u/4317177/blog/3433776