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package class49;
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// 这道题在leetcode上，所有题解都只能做到O( (logN) 平方)的解
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// 我们课上讲的是O(logN)的解
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// 打败所有题解
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public class Problem_0440_KthSmallestInLexicographicalOrder {
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public static int[] offset = { 0, 1, 10, 100, 1000, 10000, 100000, 1000000, 10000000, 100000000, 1000000000 };
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public static int[] number = { 0, 1, 11, 111, 1111, 11111, 111111, 1111111, 11111111, 111111111, 1111111111 };
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public static int findKthNumber(int n, int k) {
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// 数字num，有几位，len位
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// 65237, 5位，len = 5
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int len = len(n);
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// 65237, 开头数字，6，first
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int first = n / offset[len];
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// 65237，左边有几个？
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int left = (first - 1) * number[len];
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int pick = 0;
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int already = 0;
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if (k <= left) {
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// k / a 向上取整-> (k + a - 1) / a
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pick = (k + number[len] - 1) / number[len];
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already = (pick - 1) * number[len];
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return kth((pick + 1) * offset[len] - 1, len, k - already);
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}
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int mid = number[len - 1] + (n % offset[len]) + 1;
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if (k - left <= mid) {
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return kth(n, len, k - left);
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}
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k -= left + mid;
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len--;
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pick = (k + number[len] - 1) / number[len] + first;
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already = (pick - first - 1) * number[len];
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return kth((pick + 1) * offset[len] - 1, len, k - already);
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}
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public static int len(int n) {
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int len = 0;
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while (n != 0) {
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n /= 10;
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len++;
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}
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return len;
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}
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public static int kth(int max, int len, int kth) {
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// 中间范围还管不管的着！
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// 有任何一步，中间位置没命中，左或者右命中了，那以后就都管不着了！
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// 但是开始时，肯定是管的着的！
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boolean closeToMax = true;
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int ans = max / offset[len];
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while (--kth > 0) {
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max %= offset[len--];
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int pick = 0;
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if (!closeToMax) {
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pick = (kth - 1) / number[len];
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ans = ans * 10 + pick;
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kth -= pick * number[len];
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} else {
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int first = max / offset[len];
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int left = first * number[len];
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if (kth <= left) {
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closeToMax = false;
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pick = (kth - 1) / number[len];
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ans = ans * 10 + pick;
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kth -= pick * number[len];
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continue;
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}
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kth -= left;
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int mid = number[len - 1] + (max % offset[len]) + 1;
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if (kth <= mid) {
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ans = ans * 10 + first;
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continue;
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}
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closeToMax = false;
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kth -= mid;
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len--;
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pick = (kth + number[len] - 1) / number[len] + first;
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ans = ans * 10 + pick;
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kth -= (pick - first - 1) * number[len];
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}
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}
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return ans;
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}
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}
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