LeetCode 第 17 题 Java 实现
电话号码的字母组合
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65class Solution {
static Map<String, String> map;
static {
map = new HashMap<String, String>();
map.put("0", "");
map.put("1", "");
map.put("2", "abc");
map.put("3", "def");
map.put("4", "ghi");
map.put("5", "jkl");
map.put("6", "mno");
map.put("7", "pqrs");
map.put("8", "tuv");
map.put("9", "wxyz");
}
public List<String> letterCombinations(String digits) {
if(digits == null || digits.length() == 0){
return new ArrayList<String>();
}
return recusiveResult(digits, 0, digits.length() - 1);
}
private List<String> recusiveResult(String digits, int left, int right) {
List<String> result = new ArrayList<>();
if (left == right) {
String str = map.get(digits.charAt(left) + "");
for (int j = 0; j < str.length(); j++) {
result.add(str.charAt(j) + "");
}
} else if (left == right - 1) {
String Sleft = map.get(digits.charAt(left) + "");
String Sright = map.get(digits.charAt(right) + "");
if (Sleft != null && Sleft.length() > 0 && Sright != null && Sright.length() > 0) {
for (int i = 0; i < Sleft.length(); i++) {
for (int j = 0; j < Sright.length(); j++) {
result.add(Sleft.charAt(i) + "" + Sright.charAt(j));
}
}
} else {
if (Sleft != null && Sleft.length() > 0) {
for (int i = 0; i < Sleft.length(); i++) {
result.add(Sleft.charAt(i) + "");
}
} else {
for (int i = 0; i < Sright.length(); i++) {
result.add(Sright.charAt(i) + "");
}
}
}
} else {
int mid = (left + right) / 2;
List<String> lefStrings = recusiveResult(digits, left, mid);
List<String> rightStrings = recusiveResult(digits, mid + 1, right);
for (int i = 0; i < lefStrings.size(); i++) {
for (int j = 0; j < rightStrings.size(); j++) {
result.add(lefStrings.get(i) + rightStrings.get(j));
}
}
}
return result;
}
}
