LeetCode 34. Search for a Range

最新推荐文章于 2024-03-22 00:51:13 发布

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最新推荐文章于 2024-03-22 00:51:13 发布

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分类专栏： LeetCode 题解 Multiple Passes

本文链接：https://blog.youkuaiyun.com/github_34333284/article/details/51258172

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本文介绍了一种算法，用于在一个已排序的整数数组中找到给定目标值的起始和结束位置，确保运行时间为O(log n)。通过三种不同的实现方式展示了如何高效地解决这一问题。

Second solution is my favourite actually. But it seems that some interviewer prefer to do it recursively.

#include <iostream>
#include <vector>
using namespace std;

/*
    Given a sorted arry of integers. Find the starting and ending position
    of a given target value.
    algorithm's run time should be in order of O(lgn)
    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];

*/
// First solution. O(lgn) but worst time O(n).
vector<int> searchRange(vector<int>& nums, int target) {
    int left = 0, right = nums.size() - 1;
    vector<int> res;
    while(left <= right) {
        int mid = (left + right) / 2;
        if(nums[mid] == target) {
            int left = mid;
            int right = mid;
            while(nums[left] == target) {left--;}
            while(nums[right] == target) {right++;}
            if((nums[left+1] == target) && (nums[right-1] == target)) {
                res.push_back(left+1);
                res.push_back(right-1);
                return res;
            }
        } else if(nums[mid] < target) left = mid + 1;
        else right = mid - 1;
    }
    return {-1, -1};
}

int searchRangeRight(vector<int>& nums, int target) {
    int left = 0, right = nums.size() - 1;
    while(left < right) {
        int mid = (left + right + 1) / 2; // here should plus 1, make it moving.
        if(nums[mid] > target) right = mid - 1;
        else left = mid;
    }
    if(nums[left] == target) return left;
    return -1;
}

int searchRangeLeft(vector<int>& nums, int target) {
    int left = 0, right = nums.size() - 1;
    while(left < right) {
        int mid = (left + right) / 2;
        if(nums[mid] < target) left = mid + 1;
        else right = mid;
    }
    if(nums[right] == target) return right;
    return -1;
}

// Second solution. always O(lgn) even with repeations.
vector<int> searchRangeII(vector<int>& nums, int target) {
    int left = 0, right = nums.size() - 1;
    int leftIndex = searchRangeLeft(nums, target);
    int rightIndex = searchRangeRight(nums, target);
    return {leftIndex, rightIndex};
}

int searchRangeIII(vector<int>& nums, int left, int right, int target, bool flag) {
    if(left > right) return -1;
    int mid = (left + right) / 2;
    if(nums[mid] == target) {
        int pos = flag ? searchRangeIII(nums, left, mid - 1, target, flag) : searchRangeIII(nums, mid + 1, right, target, flag);
        return pos == -1 ? mid : pos;
    } else if(nums[mid] < target)
        searchRangeIII(nums, mid + 1, right, target, flag);
    else
        searchRangeIII(nums, left, mid - 1, target, flag);
}

// Let's do it recursively....
vector<int> searchRangeIII(vector<int>& nums, int target) {
    int left = 0, right = nums.size() - 1;
    bool leftFlag = true;
    int leftIndex = searchRangeIII(nums, left, right, target, leftFlag);
    int rightIndex = searchRangeIII(nums, left, right, target, !leftFlag);
    return {leftIndex, rightIndex};
}


int main(void) {
    vector<int> nums{5, 7, 7, 8, 8, 10};
    vector<int> res = searchRangeIII(nums, 7);
    for(int i = 0; i < res.size(); ++i) {
        cout << res[i] << endl;
    }
}