lintcode： search for a range 搜索区间

题目

搜索区间

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)

解题

正常解法是二分法分别找左右边界的，时间复杂度O(logN),但是要注意边界，边界很多种情况的。

找左边界：

1.起始位置是0的单独考虑：nums[0] == target  return 0

2.正常的二分查找思想

    　　　while(left <= right){
            int median = (left + right)/2;
            if(nums[median] < target){
                left = median+1;
            }else if(nums[median]>target){
                right = median -1;
            }else if(nums[median]==target){
                if(nums[median-1]!=target)
                    return median;
                else
                    right = median - 1;
            }
        }

找右边界：

1.结束位置单独考虑：nums[right] == target  return right

2.正常的二分查找思想

while(left <= right){
            int median = (left + right)/2;
            if(nums[median] < target){
                left = median+1;
            }else if(nums[median]>target){
                right = median -1;
            }else if(nums[median]==target){
                if(nums[median+1]!=target)
                    return median;
                else
                    left = median + 1;
            }
        }

Java

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) {
        // write your code here
        int[] res = {-1,-1};
        int len = A.length;
        if(len == 0 || A == null)
            return res;
        res[0] = BinaryLeft(A,0,len-1,target);
        res[1] = BinaryRight(A,0,len-1,target);
        return res;
    }
    public int BinaryLeft(int[] nums,int left,int right,int target){
        if(nums[0] == target)
            return 0;
        while(left <= right){
            int median = (left + right)/2;
            if(nums[median] < target){
                left = median+1;
            }else if(nums[median]>target){
                right = median -1;
            }else if(nums[median]==target){
                if(nums[median-1]!=target)
                    return median;
                else
                    right = median - 1;
            }
        }
        return -1;
    }

    public int BinaryRight(int[] nums,int left,int right,int target){
        if(nums[right] == target)
            return right;
        while(left <= right){
            int median = (left + right)/2;
            if(nums[median] < target){
                left = median+1;
            }else if(nums[median]>target){
                right = median -1;
            }else if(nums[median]==target){
                if(nums[median+1]!=target)
                    return median;
                else
                    left = median + 1;
            }
        }
     return -1;
    }

}

Java Code

Python

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):
        # write your code here
        res = [-1,-1]
        if A==None or len(A) == 0:
            return res
        res[0] = self.BoundaryLeft(A,0,len(A)-1,target)
        res[1] = self.BoundaryRight(A,0,len(A)-1,target)
        return res
    def BoundaryLeft(self,A,left,right,target):
        if A[0] == target:
            return 0
        while left<= right:
            median = (left + right)/2
            if A[median] < target:
                left = median + 1
            elif A[median] > target:
                right = median -1
            elif A[median] == target:
                if A[median-1]!=target:
                    return median
                else:
                    right = median -1
        return -1

    def BoundaryRight(self,A,left,right,target):
        if A[right] == target:
            return right
        while left<= right:
            median = (left+ right)/2
            if A[median] < target:
                left = median + 1
            elif A[median] > target:
                right = median - 1
            elif A[median] == target:
                if A[median + 1]!=target:
                    return median
                else:
                    left = median + 1
        return -1

Python Code