给定一个只包含小写字母的有序数组letters
和一个目标字母 target
,寻找有序数组里面比目标字母大的最小字母。
数组里字母的顺序是循环的。举个例子,如果目标字母target = 'z'
并且有序数组为 letters = ['a', 'b']
,则答案返回 'a'
。
示例:
输入: letters = ["c", "f", "j"] target = "a" 输出: "c" 输入: letters = ["c", "f", "j"] target = "c" 输出: "f" 输入: letters = ["c", "f", "j"] target = "d" 输出: "f" 输入: letters = ["c", "f", "j"] target = "g" 输出: "j" 输入: letters = ["c", "f", "j"] target = "j" 输出: "c" 输入: letters = ["c", "f", "j"] target = "k" 输出: "c"
注:
letters
长度范围在[2, 10000]
区间内。letters
仅由小写字母组成,最少包含两个不同的字母。- 目标字母
target
是一个小写字母。
Review:
遍历一遍就可以了,如果你闲的没事可以使用二分查找,毕竟可以快一些
ps:我很不明白一个是while(true) 一个是while(l<r) 居然是后者更快一点
同样都是遍历相同次数,居然后面的快,我搞不懂,一会反编译看一下
Code1:
while(true) 个人觉得更好一点
public char nextGreatestLetter(char[] letters, char target) {
int len = letters.length;
if (target >= letters[len - 1] || target < letters[0]) return letters[0];
int l = 0,r = len;
while (true) {
int p = (r + l) >> 1;
char c = letters[p];
if (target >= c) {
l = p + 1;
} else if (letters[p - 1] <= target) {
return c;
} else {
r = p - 1;
}
}
}
Code2:leetcode测试会快一点
class Solution {
public char nextGreatestLetter(final char[] letters, char target) {
int len = letters.length;
if (target>=letters[len-1]||target<letters[0]) return letters[0];
int r = len,l = 0;
while (l<=r){
int p =(r+l)>>1;
char c = letters[p];
if (target>=c){
l = p+1;
}else if (letters[p-1]<=target){
return c;
}else {
r = p-1;
}
}
return 'c';
}
}