Search for a Range -- Leetcode

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12.26 2014

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 begin=0;        int end=n-1;        int pos=-1;        vector<int> range;        range.push_back(-1);        range.push_back(-1);        while(begin<=end){            int mid=(begin+end)/2;            if(target==A[mid]){                pos=mid;                break;            }            else if(target>A[mid])                begin=mid+1;            else                end=mid-1;        }        if(pos==-1)            return range;        int index=pos;        while(A[index]==target && index>=0){            index--;        }        range[0]=index+1;        index=pos;        while(A[index]==target && index<=n-1){            index++;        }        range[1]=index-1;        return range;    }};

总结：

先二分查找，找到其中一个target以后，向前后遍历寻找相同的值。最后前后组成range

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