题面


解法：与leetcode321不一样的是，那个是最大值，这个是最小值，且需要处理0首位问题。那道题是Hard题中的Hard，无论是思维量还是代码量都是顶级，这道题难度并不弱于那道。
观察到数据量为500级别，可以考虑N^3的解法。那么显然便是动态规划。定义数组dp[i][j][k]为第一个数组枚举到i，第二个数组枚举到j，总共选了k个的最小答案。注意由于是数组对比大小，需要自定义比较函数。

import java.util.*;

public class Solution {
    public int[] minNumber(int[] nums1, int[] nums2, int k) {
        int m = nums1.length, n = nums2.length;
        String[][][] dp = new String[m + 1][n + 1][k + 1];
        for (int i = 0; i <= m; ++i) {
            for (int j = 0; j <= n; ++j) {
                for (int l = 0; l <= k; ++l) {
                    dp[i][j][l] = "";
                }
            }
        }

        dp[1][0][1] = Integer.toString(nums1[0]);
        dp[0][1][1] = Integer.toString(nums2[0]);

        for (int i = 2; i <= m; ++i) {
            dp[i][0][1] = min(dp[i-1][0][1], Integer.toString(nums1[i-1]));
        }

        for (int j = 2; j <= n; ++j) {
            dp[0][j][1] = min(dp[0][j-1][1], Integer.toString(nums2[j-1]));
        }

        for (int i = 1; i <= m; ++i) {
            for (int j = 1; j <= n; ++j) {
                dp[i][j][1] = min(dp[i-1][j-1][1], min(Integer.toString(nums1[i-1]), Integer.toString(nums2[j-1])));
            }
        }

        for (int l = 2; l <= k; ++l) {
            for (int i = 0; i <= m; ++i) {
                for (int j = Math.max(l - i, 0); j <= n; ++j) {
                    dp[i][j][l] = String.join("", Collections.nCopies(l, "9"));
                    if (i > 0 && !dp[i-1][j][l].isEmpty()) {
                        dp[i][j][l] = min(dp[i][j][l], dp[i-1][j][l]);
                    }
                    if (j > 0 && !dp[i][j-1][l].isEmpty()) {
                        dp[i][j][l] = min(dp[i][j][l], dp[i][j-1][l]);
                    }
                    if (j > 0 && !dp[i][j-1][l-1].isEmpty()) {
                        dp[i][j][l] = min(dp[i][j][l], dp[i][j-1][l-1] + Integer.toString(nums2[j-1]));
                    }
                    if (i > 0 && !dp[i-1][j][l-1].isEmpty()) {
                        dp[i][j][l] = min(dp[i][j][l], dp[i-1][j][l-1] + Integer.toString(nums1[i-1]));
                    }
                }
            }
        }

        int[] ret = new int[k];
        for (int i = 0; i < k; ++i) {
            ret[i] = dp[m][n][k].charAt(i) - '0';
        }

        return ret;
    }

    private String min(String a, String b) {
        if(a.length() == 0) return a;
        if(b.length() == 0) return b;
        if(a.charAt(0) == '0') a = String.join("", Collections.nCopies(500, "9"));
        if(b.charAt(0) == '0') b = String.join("", Collections.nCopies(500, "9"));
        return a.compareTo(b) < 0 ? a : b;
    }
}