61.Search for a Range-搜索区间（中等题）_搜索区间题目-优快云博客

本文介绍了一种在已排序的整数数组中查找特定目标值起始和结束位置的方法。通过两种二分查找策略实现高效搜索：一是直接查找目标值后向两侧扩展；二是分别确定目标值的下界和上界。这两种方法均可达到O(logn)的时间复杂度。

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搜索区间

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]
样例

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

时间复杂度 O(log n)
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;
    }
}