61.Search for a Range-搜索区间（中等题）

搜索区间

1. 题目

   给定一个包含 n 个整数的排序数组，找出给定目标值 target 的起始和结束位置。
   如果目标值不在数组中，则返回[-1, -1]

2. 样例

   给出[5, 7, 7, 8, 8, 10]和目标值target=8,
   返回[3, 4]

3. 挑战

   时间复杂度 O(log n)

4. 题解

1.二分法找到target，再向两边查找边界。

public class Solution {
    /** 
     *@param A : an integer sorted array
     *@param target :  an integer to be inserted
     *return : a list of length 2, [index1, index2]
     */
    public int[] searchRange(int[] A, int target) {
        if (A.length == 1)
        {
            return A[0] == target ? new int[]{0,0} : new int[]{-1,-1};
        }
        return doSearch(A,0,A.length-1,target);
    }

    private int[] doSearch(int[] A, int start, int end, int target)
    {
        while (start <= end)
        {
            int mid = start + (end - start) / 2;
            if (A[mid] == target)
            {
                int left=mid;
                int right=mid;
                while (left > 0 && A[left-1] == target)
                {
                    left--;
                }
                while (right < A.length-1 && A[right+1] == target)
                {
                    right++;
                }
                return new int[]{left,right};
            }
            else if (A[mid] < target)
            {
                start = mid + 1;
            }
            else
            {
                end = mid - 1;
            }
        }

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

2.二分法分别查找下界和上界

题解链接

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]; 

        // 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 {
            bound[0] = bound[1] = -1;
            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 {
            bound[0] = bound[1] = -1;
            return bound;
        }

        return bound;
    }
}

Last Update 2016.10.6