【leetcode】Search for a Range（middle）

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].

 

思路：

我自己用了个最简单的，普通的二分查找，然后两边延伸找起始和结束点。

int *searchRange(int A[], int n, int target) {
    int ans[2] = {-1, -1};
    int l = 0, r = n - 1;
    while(l <= r)
    {
        int m = (l + r) / 2;
        if(A[m] == target)
        {
            ans[0] = ans[1] = m;
            while(ans[0] - 1 >= 0 && A[ans[0]] == A[ans[0] - 1]) ans[0]--;
            while(ans[1] + 1 < n && A[ans[1]] == A[ans[1] + 1]) ans[1]++;
            return ans;
        }
        else if(A[m] > target)
            r = m - 1;
        else
            l = m + 1;
    }
    return ans;
}

 

大神分成了分别用两次二分查找找起始和结束位置

https://leetcode.com/discuss/18242/clean-iterative-solution-binary-searches-with-explanation中有详细解释

vector<int> searchRange(int A[], int n, int target) {
    int i = 0, j = n - 1;
    vector<int> ret(2, -1);
    // Search for the left one
    while (i < j)
    {
        int mid = (i + j) /2;
        if (A[mid] < target) i = mid + 1;
        else j = mid;
    }
    if (A[i]!=target) return ret;
    else ret[0] = i;

    // Search for the right one
    j = n-1;  // We don't have to set i to 0 the second time.
    while (i < j)
    {
        int mid = (i + j) /2 + 1;   // Make mid biased to the right
        if (A[mid] > target) j = mid - 1;  
        else i = mid;               // So that this won't make the search range stuck.
    }
    ret[1] = j;
    return ret; 
}

 

另一种通过加减0.5来找位置的方法

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, arr, target):
    start = self.binary_search(arr, target-0.5)
    if arr[start] != target:
        return [-1, -1]
    arr.append(0)
    end = self.binary_search(arr, target+0.5)-1
    return [start, end]

def binary_search(self, arr, target):
    start, end = 0, len(arr)-1
    while start < end:
        mid = (start+end)//2
        if target < arr[mid]:
            end = mid
        else:
            start = mid+1
    return start