You are given a string num, which represents a large integer. You are also given a 0-indexed integer array change of length 10 that maps each digit 0-9 to another digit. More formally, digit d maps to digit change[d].
You may choose to mutate any substring of num. To mutate a substring, replace each digit num[i] with the digit it maps to in change (i.e. replace num[i] with change[num[i]]).
Return a string representing the largest possible integer after mutating (or choosing not to) any substring of num.
A substring is a contiguous sequence of characters within the string.
Example 1:
Input: num = "132", change = [9,8,5,0,3,6,4,2,6,8] Output: "832" Explanation: Replace the substring "1": - 1 maps to change[1] = 8. Thus, "132" becomes "832". "832" is the largest number that can be created, so return it.
Example 2:
Input: num = "021", change = [9,4,3,5,7,2,1,9,0,6] Output: "934" Explanation: Replace the substring "021": - 0 maps to change[0] = 9. - 2 maps to change[2] = 3. - 1 maps to change[1] = 4. Thus, "021" becomes "934". "934" is the largest number that can be created, so return it.
Example 3:
Input: num = "5", change = [1,4,7,5,3,2,5,6,9,4] Output: "5" Explanation: "5" is already the largest number that can be created, so return it.
Constraints:
1 <= num.length <= 105numconsists of only digits0-9.change.length == 100 <= change[d] <= 9
子字符串突变后可能得到的最大整数。
给你一个字符串 num ，该字符串表示一个大整数。另给你一个长度为 10 且 下标从 0 开始 的整数数组 change ，该数组将 0-9 中的每个数字映射到另一个数字。更规范的说法是，数字 d 映射为数字 change[d] 。
你可以选择 突变 num 的任一子字符串。突变 子字符串意味着将每位数字 num[i] 替换为该数字在 change 中的映射（也就是说，将 num[i] 替换为 change[num[i]]）。
请你找出在对 num 的任一子字符串执行突变操作（也可以不执行）后，可能得到的 最大整数 ，并用字符串表示返回。
子字符串 是字符串中的一个连续序列。
来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/largest-number-after-mutating-substring
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这道题的思路是贪心。题目问是否有可能根据change数组的情况，合理替换某一段数字（子串）从而使得替换之后的数字比原来的数字更大。所以为了能使结果更大，我们需要试图从 input 字符串的最左边，也就是数字的最高位开始试图替换。如果能把某一位数字替换得更大，则替换，并用一个 flag 记录我们已经开始替换了。如果这一路下去一直可以替换一个更大的数字或者起码保持不变，那么就一直走下去；如果已经开始替换了但是中间遇到一个地方只能换成一个小的数字，则在这里停下。已经置换的部分即是题意要求的子串。返回替换后的结果即可。
时间O(n)
空间O(n)
Java实现
1 class Solution { 2 public String maximumNumber(String num, int[] change) { 3 char[] chars = num.toCharArray(); 4 boolean changed = false; 5 for (int i = 0; i < chars.length; i++) { 6 int cur = chars[i] - '0'; 7 int candidate = change[cur]; 8 if (candidate > cur) { 9 chars[i] = (char) (candidate + '0'); 10 changed = true; 11 } 12 if (candidate < cur && changed) { 13 break; 14 } 15 } 16 return new String(chars); 17 } 18 }