注：无思路 参考答案
code
class Solution {
static final int N = 23;
static final int M = 401;
static final int MOD = 1000000007;
int[][] d;
String num;
int min_sum;
int max_sum;
public int count(String num1, String num2, int min_sum, int max_sum) {
d = new int[N][M];
for (int i = 0; i < N; i++) {
Arrays.fill(d[i], -1);
}
this.min_sum = min_sum;
this.max_sum = max_sum;
return (get(num2) - get(sub(num1)) + MOD) % MOD;
}
public int get(String num) {
this.num = new StringBuffer(num).reverse().toString();
return dfs(num.length() - 1, 0, true);
}
// 求解 num - 1，先把最后一个非 0 字符减去 1，再把后面的 0 字符变为 9
public String sub(String num) {
char[] arr = num.toCharArray();
int i = arr.length - 1;
while (arr[i] == '0') {
i--;
}
arr[i]--;
i++;
while (i < arr.length) {
arr[i] = '9';
i++;
}
return new String(arr);
}
public int dfs(int i, int j, boolean limit) {
if (j > max_sum) {
return 0;
}
if (i == -1) {
return j >= min_sum ? 1 : 0;
}
if (!limit && d[i][j] != -1) {
return d[i][j];
}
int res = 0;
int up = limit ? num.charAt(i) - '0' : 9;
for (int x = 0; x <= up; x++) {
res = (res + dfs(i - 1, j + x, limit && x == up)) % MOD;
}
if (!limit) {
d[i][j] = res;
}
return res;
}
}
作者：力扣官方
原文链接: https://blog.csdn.net/qq_53568730/article/details/135636218