Search for a Range 分类： Leetcode（查找） Leetcode（排序） 2015-04-10 15:34 23人阅读 评论(0) 收藏

Search for a Range

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].

 

class Solution {
public:
    vector<int> searchRange(int A[], int n, int target) {
        int i = 0, j = n-1;
        vector<int> vec;
        while( A[i] != target && i < n ) {
            i++;
        }
        while( A[j] != target && j > 0 ) {
            j--;
        }
        
        if( i == n  || j== -1) {
            vec.push_back(-1);
            vec.push_back(-1);
        }
        
        else {
            vec.push_back(i);
            vec.push_back(j);
        }
        
        return vec;
    }
};

class Solution {
public:
    vector<int> searchRange(int A[], int n, int target) {
        int i = 0, j = n-1;
        vector<int> ret(2,-1);
        
        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;
        
        j = n-1;
        while(i < j) {
            int mid = (i+j+1) /2 ;
            if (A[mid] > target) j = mid-1;
            else i = mid;
        }
        ret[1] = j;
        return ret;
    }
};

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