试题
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”
代码
1、如果等于的话有两种可能:一是正好右边就是最小元素,此时left+1刚好,另外就是右边元素是target的重复值,此时left+1相当于区间的更新。
2、如果target大于mid,直接更新区间即可
3、如果target小于mid,一种极端情况就是此时mid刚好是最小元素,区间更新后最小元素将会出现在区间之外,这时候循环终止条件必须有等于才能使得left+1能够定位到最小元素。
class Solution {
public char nextGreatestLetter(char[] letters, char target) {
int left=0, right=letters.length-1;
while(left<=right){
int mid = left+(right-left)/2;
if(letters[mid]==target){
left = mid+1;
}else if(letters[mid]>target){
right = mid-1;
}else if(letters[mid]<target){
left = mid+1;
}
}
return left<letters.length?letters[left]:letters[0];
}
}