Find Smallest Letter Greater Than Target【找到比目标数大的数字中的最小数】【二分的终止判断条件】

PROBLEM:

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:

1. letters has a length in range [2, 10000].
2. letters consists of lowercase letters, and contains at least 2 unique letters.
3. target is a lowercase letter.

SOLVE:

class Solution {
public:
    char nextGreatestLetter(vector& letters, char target) {
        if (letters.back() <= target) return letters.front();
        int low = 0, high = letters.size() - 1;
        while (low < high) {
            auto mid = (low + high) / 2;
            if (target < letters[mid]) high = mid;
            else low = mid + 1;
        }
        return letters[low];
    }
};

分析：用二分查找效率是最高的。这里需要注意：当target==mid时，和target>mid时是一样的，所需要的输出至少都是下一个；其次还要注意这里可能会有多个值重复，所以不能在target==mid时直接输出下一个字符，很可能还是target！

总结：定义mid=start+(end-start)/2（如果写成mid=(end+start)/2的话可能会发生溢出）的情况下，如果缩小区间表达式为{end=mid-1//start=mid+1}，最后的终止判定条件是start>=end；如果缩小区间表达式为{end=mid//start=mid+1}（{end=mid-1//start=mid}不行，假设），最后的终止判定条件是start==end，当然start>=end也没错；如果缩小区间表达式为{end=mid//start=mid}，最后的终止判定条件为start=end-1。