題目描述

Given an array of integers sorted in ascending order, 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

用二分查詢尋找等於目標值的下標，找到後向前向後搜尋，找到目標值的起始索引和結束索引值。

C++實現

class Solution {
public:
    vector<int> searchRange(vector<int>& nums, int target) {
        vector<int> ret;
        ret.push_back(-1);
        ret.push_back(-1);

        int l,r,m=0,t=-1,i,pm=-1;
        l=0;
        r=nums.size()-1
 
;
        while(l<r && nums[l]<target && nums[r]>target && r!=1)
        {
            m=(l+r)/2;
            if(m==pm)
                break;
            pm=m;
            if(nums[m]<target)
                l=m;
            else if(nums[m]>target)
                r=m;
            else
 
 break;
        }

        if(nums.size()>0)
        {
            if(nums[l]==target)
                t=l;
            if(nums[r]==target)
                t=r;
            if(nums[m]==target)
                t=m;

            if(t>=0)
            {
                for(i=t;nums[i]==target && i>=0;i--);
                ret[0]=i+1;
                for(i=t;nums[i]==target && i<nums.size();i++);
                ret[1]=i-1;
            }   
        }

        return ret;
    }
};

C實現

/**
 * Return an array of size *returnSize.
 * Note: The returned array must be malloced, assume caller calls free().
 */
int* searchRange(int* nums, int numsSize, int target, int* returnSize) {
    int* ret=(int*)malloc(2*sizeof(int));
    ret[0]=-1;
    ret[1]=-1;
    *returnSize=2;
    if(numsSize==0)
        return ret;

    int l,r,m=0;
    int t=-1;
    int i;
    l=0;
    r=numsSize-1;
    while(l<r && nums[l]<target && nums[r]>target)
    {
        m=(l+r)/2;
        if(nums[m]<target)
            l=m+1;
        else if(nums[m]>target)
            r=m-1;
        else
            break;
    }

    if(nums[l]==target)
        t=l;
    if(nums[r]==target)
        t=r;
    if(nums[m]==target)
        t=m;

    if(t>=0)
    {
        for(i=t;nums[i]==target && i>=0;i--);
        ret[0]=i+1;
        for(i=t;nums[i]==target && i<numsSize;i++);
        ret[1]=i-1;
    }

    return ret;
}