Search for a Range

最新推荐文章于 2022-03-08 23:18:06 发布

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本文介绍了一种在已排序数组中查找指定目标值起始和结束位置的方法，通过两次二分查找实现O(logn)的时间复杂度。示例：对于数组[5,7,7,8,8,10]及目标值8，返回[3,4]。

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

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

Have you met this question in a real interview? Yes
Example
Given [5, 7, 7, 8, 8, 10] and target value 8,
return [3, 4].

Challenge
O(log n) time.

两次二分查找!!!条件不一样..找左边界尽量向左,找右边界尽量向右!!!

public class Solution {
    public int[] searchRange(int[] A, int target) {
        if (A.length == 0) {
            return new int[]{-1, -1};
        }

        int start, end, mid;
        int[] bound = new int[2]; 
        bound[0]=-1;
        bound[1]=-1;
        // search for left bound
        start = 0; 
        end = A.length - 1;
        while (start + 1 < end) {
            mid = start + (end - start) / 2;
            if (A[mid] == target) {
                end = mid;
            } else if (A[mid] < target) {
                start = mid;
            } else {
                end = mid;
            }
        }
        if (A[start] == target) {
            bound[0] = start;
        } else if (A[end] == target) {
            bound[0] = end;
        } else {

            return bound;
        }

        // search for right bound
        start = 0;
        end = A.length - 1;
        while (start + 1 < end) {
            mid = start + (end - start) / 2;
            if (A[mid] == target) {
                start = mid;
            } else if (A[mid] < target) {
                start = mid;
            } else {
                end = mid;
            }
        }
        if (A[end] == target) {
            bound[1] = end;
        } else if (A[start] == target) {
            bound[1] = start;
        } else {

            return bound;
        }

        return bound;
    }
}