[leetcode] Search for a Range

Given a sorted array of integers, find the starting and ending position of a given target value.

Your algorithm's runtime complexity must be in the order of O(log n).

If the target is not found in the array, return[-1, -1].

For example,
Given[5, 7, 7, 8, 8, 10]and target value 8,
return[3, 4].

https://oj.leetcode.com/problems/search-for-a-range/

排过序的数组，要求时间是logn，一看就是要用binary search。 值得一读， 值得一读2

思路1：binary search到元素，然后向两边扩展找相同元素，最坏情况O(n).

思路2：对binary search稍微修改，然后进行两次binary search，分别定位指定元素的下界和上界。

思路2代码：

已删除

 第二遍记录：

　　两个函数的名字改一下分别叫：lowerBound 和 upperBound。

public class Solution {
    public int[] searchRange(int[] A, int target) {
        if (A == null || A.length == 0)
            return new int[] { -1, -1 };
        int n = A.length;
        int floor = lowerBound(A, target);
        int ceiling = upperBound(A, target);
        if (ceiling > floor)
            return new int[] { floor, ceiling - 1 };

        else
            return new int[] { -1, -1 };
    }

    public int lowerBound(int[] a, int x) {
        int left = 0;
        int right = a.length - 1;

        while (left <= right) {
            int mid = left + (right - left) / 2;
            if (x > a[mid]) {
                left = mid + 1;
            } else if (x <= a[mid]) {
                right = mid - 1;
            }
        }
        return left;
    }

    public int upperBound(int[] a, int x) {
        int left = 0;
        int right = a.length - 1;

        while (left <= right) {
            int mid = left + (right - left) / 2;
            if (x >= a[mid]) {
                left = mid + 1;
            } else if (x < a[mid]) {
                right = mid - 1;
            }
        }
        return left;
    }

}

第N遍，

分析为什么left是最终需要的坐标，考虑最后left和right相交的两种情况（lrm都在最后一个元素，lrm在最后两个元素），分情况讨论。