题目大意

查找升序数组第一次出现target数字的范围，返回索引号。题目要求的时间复杂度是O(log n)。

解题思路

二分查找变种，二分法时间复杂度就是O(log n)

代码

Java: 重复数组中的二分法找最左

https://github.com/CyC2018/Interview-Notebook/blob/master/notes/Leetcode%20%E9%A2%98%E8%A7%A3.md#%E4%BA%8C%E5%88%86%E6%9F%A5%E6%89%BE

public int[] searchRange(int[] nums, int target) {
    int first = binarySearch(nums, target);
    int last = binarySearch(nums, target + 1) - 1;
    if (first == nums.length || nums[first] != target)
        return new int[]{-1, -1};
    else
        return new int[]{first, Math.max(first, last)};
}

private int binarySearch(int[] nums, int target) {
    int l = 0, h = nums.length; // 注意 h 的初始值
    while (l < h) {
        int m = l + (h - l) / 2;
        if (nums[m] >= target)
            h = m;
        else
            l = m + 1;
    }
    return l;
}

Python：二分法找到后直接往前后遍历

class Solution:
    # @param A, a list of integers
    # @param target, an integer to be searched
    # @return a list of length 2, [index1, index2]
    def searchRange(self, A, target):
        left = 0; right = len(A) - 1
        while left <= right:
            mid = (left + right) / 2
            if A[mid] > target:
                right = mid - 1
            elif A[mid] < target:
                left = mid + 1
            else:
                list = [0, 0]
                if A[left] == target: list[0] = left
                if A[right] == target: list[1] = right
                for i in range(mid, right+1):
                    if A[i] != target: list[1] = i - 1; break
                for i in range(mid, left-1, -1):
                    if A[i] != target: list[0] = i + 1; break
                return list
        return [-1, -1]

总结