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package class26;
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import java.util.LinkedList;
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import java.util.List;
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// 本题测试链接 : https://leetcode.com/problems/expression-add-operators/
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public class Code03_ExpressionAddOperators {
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public static List<String> addOperators(String num, int target) {
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List<String> ret = new LinkedList<>();
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if (num.length() == 0) {
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return ret;
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}
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// 沿途的数字拷贝和+ - * 的决定，放在path里
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char[] path = new char[num.length() * 2 - 1];
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// num -> char[]
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char[] digits = num.toCharArray();
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long n = 0;
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for (int i = 0; i < digits.length; i++) { // 尝试0~i前缀作为第一部分
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n = n * 10 + digits[i] - '0';
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path[i] = digits[i];
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dfs(ret, path, i + 1, 0, n, digits, i + 1, target); // 后续过程
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if (n == 0) {
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break;
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}
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}
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return ret;
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}
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// char[] digits 固定参数，字符类型数组，等同于num
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// int target 目标
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// char[] path 之前做的决定，已经从左往右依次填写的字符在其中，可能含有'0'~'9' 与 * - +
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// int len path[0..len-1]已经填写好，len是终止
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// int pos 字符类型数组num, 使用到了哪
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// left -> 前面固定的部分 cur -> 前一块
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// 默认 left + cur ...
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public static void dfs(List<String> res, char[] path, int len,
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long left, long cur,
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char[] num, int index, int aim) {
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if (index == num.length) {
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if (left + cur == aim) {
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res.add(new String(path, 0, len));
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}
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return;
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}
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long n = 0;
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int j = len + 1;
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for (int i = index; i < num.length; i++) { // pos ~ i
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// 试每一个可能的前缀，作为第一个数字！
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// num[index...i] 作为第一个数字！
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n = n * 10 + num[i] - '0';
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path[j++] = num[i];
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path[len] = '+';
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dfs(res, path, j, left + cur, n, num, i + 1, aim);
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path[len] = '-';
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dfs(res, path, j, left + cur, -n, num, i + 1, aim);
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path[len] = '*';
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dfs(res, path, j, left, cur * n, num, i + 1, aim);
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if (num[index] == '0') {
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break;
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}
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}
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}
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}
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