Given a list of sorted characters letters containing only lowercase letters, and given a target letter target, find the smallest element in the list that is larger than the given target.
Letters also wrap around. For example, if the target is target = 'z' and letters = ['a', 'b'], the answer is 'a'.
Examples:
Input:
letters = ["c", "f", "j"]
target = "a"
Output: "c"
Input:
letters = ["c", "f", "j"]
target = "c"
Output: "f"
Input:
letters = ["c", "f", "j"]
target = "d"
Output: "f"
Input:
letters = ["c", "f", "j"]
target = "g"
Output: "j"
Input:
letters = ["c", "f", "j"]
target = "j"
Output: "c"
Input:
letters = ["c", "f", "j"]
target = "k"
Output: "c"
Note:
letters has a length in range [2, 10000].
letters consists of lowercase letters, and contains at least 2 unique letters.
target is a lowercase letter.
Code
- 二分法，时间复杂度O(logN)，空间复杂度O(1)
思路
和459.排序数组中最接近元素思路类似，细节上差别
class Solution {
public char nextGreatestLetter(char[] letters, char target) {
if (letters == null || letters.length == 0) {
return 'A';
}
int index = findIndex(letters, target);
if (index == 0) {
return letters[0];
}
if (index == letters.length) {
return letters[0];
}
return letters[index];
}
private int findIndex(char[] letters, char target) {
int start = 0;
int end = letters.length - 1;
while (start + 1 < end) {
int mid = start + (end - start) / 2;
if (letters[mid] <= target) {
start = mid;
} else {
end = mid;
}
}
if (letters[start] > target) {
return start;
}
if (letters[end] > target) {
return end;
}
return letters.length;
}
}
- 记录浏览过的字母，时间复杂度O(N)，空间复杂度O(1)
思路
用 seen[26] 布尔数组存储是否遍历过相应字母，从 target 对应字母开始递增查找，找到的第一个seen[target - 'a']为true对应的字母就是我们要查找的值，如果 target 大于z时还没找到，将 target 置为a, 继续查找
class Solution {
public char nextGreatestLetter(char[] letters, char target) {
boolean[] seen = new boolean[26];
for (char c: letters)
seen[c - 'a'] = true;
while (true) {
target++;
if (target > 'z') target = 'a';
if (seen[target - 'a']) return target;
}
}
}
- 遍历，时间复杂度O(N)，空间复杂度O(1)
思路
字母数组是排序的，遍历字母数组，找到大于 target 的字母，遍历到最后还没找到就返回 letters[0]
class Solution {
public char nextGreatestLetter(char[] letters, char target) {
for (char c: letters)
if (c > target) return c;
return letters[0];
}
}
后两种方法都是暴力解法
