John's business

There are n cities on an axis, numbers from 0 ~ n - 1. John intends to do business in these n cities, He is interested in Armani’s shipment. Each city has a price for these goods prices [i]. For city x, John can buy the goods from the city numbered from x - k to x + k, and sell them to city x. We want to know how much John can earn at most in each city?

注意事項prices.length range is [2, 100000], k <= 100000.

範例
Given prices = [1, 3, 2, 1, 5], k = 2, return [0, 2, 1, 0, 4].

Explanation：i = 0, John can go to the city 0 ~ 2. He can not make money because the prices in city 1 and city 2 are both higher than the price in city 0, that is, ans[0] = 0;i = 1, John can go to the city 0~3. He can buy from city 0 or city 3 to earn the largest price difference. That is, ans[1] = 2.i = 2, John can go to the city 0~4. Obviously, he can earn the largest price difference by buying from city 3. That is, ans[2] = 1.i = 3, John can go to the city 1~4. He can not make money cause city 3 has the lowest price. That is, ans[3] = 0.i = 4, John can go to the city 2~4. He can earn the largest price difference by buying from city 3. That is, ans[4] = 4.

Given prices = [1, 1, 1, 1, 1], k = 1, return [0, 0, 0, 0, 0]

Explanation:All cities are the same price, so John can not make money, that is, all ans are 0.

第一次，TLE

#ifndef C751_H#define C751_H#include<iostream>#include<vector>using namespace std;class Solution {public:    /**    * @param A: The prices [i]    * @param k:    * @return: The ans array    */    vector<int> business(vector<int> &A, int k) {        // Write your code here        int len = A.size();        vector<int> res(len, 0);        int min = INT_MAX;        for (auto c : A)        {            if (min > c)                min = c;        }        if (k >= len)        {            for (auto &c : A)                c -= min;            return A;        }        for (int i = 0; i < len; ++i)        {            int value = INT_MIN;            if (i - k <= 0 && i + k >= len - 1)            {                value = A[i] - min;            }            else if (i - k <= 0 && i + k < len - 1)            {                for (int j = 0; j <= i + k; ++j)                {                    value = maxVal(value, A[i] - A[j]);                }            }            else if (i - k>0 && i + k >= len - 1)            {                for (int j = i - k; j < len; ++j)                {                    value = maxVal(value, A[i] - A[j]);                }            }            else            {                for (int j = i - k; j <= i + k; ++j)                {                    value = maxVal(value, A[i] - A[j]);                }            }            res[i] = value;        }        return res;    }    int maxVal(int a, int b)    {        return a>b ? a : b;    }};#endif

題目本質上就是求對於位置i，找到區間[i-k,i+k]上的最小值min，將新值取A[i]-min。
可以利用線段樹：

#ifndef C751_H#define C751_H#include<iostream>#include<vector>using namespace std;class SegmentNode{public:    int start, end, min;    SegmentNode *left, *right;    SegmentNode(int start, int end, int min){        this->start = start;        this->end = end;        this->min;        this->left = this->right = NULL;    }};class Solution {public:    /**    * @param A: The prices [i]    * @param k:    * @return: The ans array    */    vector<int> business(vector<int> &A, int k) {        // Write your code here        int len = A.size();        vector<int> res(len, 0);        SegmentNode *root = build(A, 0, len - 1);        for (int i = 0; i<len; ++i)        {            int start = i - k;            int end = i + k;            //修正區間範圍            if (start < 0)                start = 0;            if (end >= len)                end = len - 1;            res[i] = A[i]-query(root, start, end);        }        return res;    }    //查詢線段樹在區間[start,end]上的最小值    int query(SegmentNode *root, int start, int end)    {        if (start > end || !root)            return 0;        SegmentNode *node = root;        if (start == node->start&&end == node->end)            return node->min;        if (start >= node->right->start)            return query(node->right, start, end);        else if (end <= node->left->end)            return query(node->left, start, end);        else        {            return minVal(query(node->left, start, node->left->end), query(node->right, node->right->start, end));        }    }    //構建線段樹,節點包含區間內的最小值    SegmentNode *build(vector<int> &A, int start, int end)    {        if (start > end)            return NULL;        SegmentNode *node = new SegmentNode(start, end, INT_MAX);        if (start == end)            node->min = A[start];        else        {            node->left = build(A, start, (start + end) / 2);            node->right = build(A, (start + end) / 2 + 1, end);            node->min = minVal(node->left->min, node->right->min);        }        return node;    }    int minVal(int a, int b)    {        return a < b ? a : b;    }};#endif