[LeetCode]Search for a Range

標籤：

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].

[LeetCode Source]

思路：

1）先用二分尋找找到點，然後再左右擴充，找到範圍。如果有m個要搜尋點，演算法複雜度為O(lgN+m),m<=N。最好的情況下是O(lgN)，只有一個點的時候；最差情況O(N)，全是要搜尋點的時候。這種做法可以Ac。

class Solution {public:    vector<int> searchRange(vector<int>& nums, int target) {        vector<int> result(2,-1);        int left = 0;        int right = nums.size()-1;        int mid = 0;        int first = -1;        int second = -1;        while(left<=right){            mid = left+(right-left)/2;            if(nums[mid] > target)                right = mid - 1;            if(nums[mid] < target)                left = mid + 1;            if(nums[mid] == target){                first = mid;                second = mid;                break;            }        }        for(int i=mid-1;i>=0;i--){            if(nums[i]==target)                first--;            else break;        }        for(int i=mid+1;i<nums.size();++i){            if(nums[i]==target)                second++;            else break;        }        result[0] = first;        result[1] = second;        return result;    }};

2）可能是最優解法。兩次二分尋找，一次是找到左邊界點，第二次找到右邊界點。

找左邊界點時，修改二分尋找的規則，如果找到點，還要繼續往左二分直到找到左邊界點；

找右邊界點時，同理。

特別注意break的處理，當明顯找不到點時break掉迴圈，對於找左邊界點就是nums[right]<target。

這種解法複雜度是O(lgN)

class Solution {public:    vector<int> searchRange(vector<int>& nums, int target) {    int hit = -1;    int left=0, right=nums.size() - 1;    int first = 0, second = 0;    vector<int> ret(2, -1);    while(left <= right)    {        int mid = left + (right - left) / 2;        if(nums[right] < target)    break;        if(nums[mid] >= target)        {            if(nums[mid] == target) hit = mid;            right = mid - 1;        }        else        {            left = mid + 1;        }    }    if(hit == -1)   return ret;    first = hit;    left = hit;    right = nums.size() - 1;    hit = -1;    while(left <= right)    {        int mid = left + (right - left) / 2;        if(nums[left] > target)    break;        if(nums[mid] <= target)        {            if(nums[mid] == target) hit = mid;            left = mid + 1;        }        else        {            right = mid - 1;        }    }    if(hit == -1)   return ret;    second = hit;    ret[0] = first;    ret[1] = second;    return ret;}};

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[LeetCode]Search for a Range