Friday, November 27, 2015

ITint5 2- 环形路加油


https://github.com/AnnieKim/ITint5/blob/master/002_%E7%8E%AF%E5%BD%A2%E8%B7%AF%E5%8A%A0%E6%B2%B9.cpp
 链接: http://www.itint5.com/oj/#2
 问题:
 有一个环形公路上有n个加油站,第i个加油站的油量为a[i]。
 假设有一辆邮箱体积无穷大的汽车,初始邮箱是空的,汽车从加油站i行驶到加油站i+1需耗油g[i]。
 问是否能够选出某个加油站作为起点,使汽车能够绕环形公路行驶一圈返回到该加油站。
 实现函数int selectGasStation(int a[], int g[], int n)。
 如果存在满足条件的加油站,返回该加油站的序号(0-based)。否则返回-1。
 提示:n可能达到106,O(n^2)的枚举算法会超出时间限制。
http://xsk.tehon.org/den/index.php/category/tech/gas-station-on-the-circle.html
方法一:从左往右遍历,在油量和最少的位置的下一个位置出发。
int res = 0, min = N[0] - g[0], sum = min, i = 1;
while(i < n) {
    sum += N[i] - g[i];
    if(sum < min) {
        min = sum;
        res = i;
    }
    ++i;
}
return sum >= 0 ? (res + 1) % n : -1;


方法二:开辟一个长度为N的数组v,记录N[i]-g[i]。从后往前遍历数组v,如果v[i]小于零,将其与v[i-1]合并,因为此时i不可能作为起点。如果v[i]不小于零,记入sum,并记录该位置pos(有可能作为起点)。最后,如果v[0]大于等于零,说明整个路段已经没有负的v[i]。返回0即可。如果v[0]+sum>=0,有满足条件的加油站,返回pos。否则,返回-1。
int v[n];
for(int i = 0; i < n; ++i)
    v[i] = N[i] - g[i];
int sum = 0, pos = -1;
for(int i = n-1; i > 0; --i) {
    if(v[i] >= 0) {
        sum += v[i];
        pos = i;
    } else {
        v[i-1] += v[i];
    }
}
if(v[0] >= 0) 
    return 0;
else if(v[0] + sum >= 0)
    return pos;
else 
    return -1;
方法三:开辟一个长度为2*n-1的数组v,记录N[i]-g[i],这样就把一个环转化为一条线。使用两个指针start和end。如果[start,end]区间和小于0,令start=end+1并继续。直至找到长度为N的区间[start,end],并且区间和大于等于0。找到返回start,找不到返回-1。
int v[2 * n];
for (int i = 0; i < n; ++i) {
    v[i] = N[i] - g[i];
    v[i+n] = N[i] - g[i];
}
int sum = 0;
for (int start = 0, end = 0; start <= n && end < 2 * n; end++) {
    if(sum + v[end] < 0) {
        start = end + 1;
        sum = 0;
    } else {
        if(end - start == n - 1) 
            return start;
        sum += v[end];
    }
}
return -1;
以上三种解法的时间复杂度均为O(n)。对于任意一个O(n)复杂度的DP题目,必有辅助数组仅与n有关,若刨除不需要计算的量还与两个以上的量有关,则不可能做到O(n)。
http://lixinzhang.github.io/itint5-mian-shi-ti-zong-jie.html
环形,即可以考虑2*N-1的展开处理。 定义c[i]=a[i]g[i]表示完成某一路段真实耗油量(因为起点加了一些油)。于是,问题转化为求以某一起点slow开始到长度n节点的所有前缀和均为正的问题。
以slow点起始位置,如果在途中的fast点,出现前缀和为负数的情况,那么slow到fast之间的点,则不可能满足约束。因此直接跳过这些点,从slow = fast +1开始找。因此,时间复杂度为O(N)
int selectGasStation(const vector<int> &a, const vector<int> &g) {
    if(a.size() == 0 || g.size() == 0 || a.size() != g.size()) return -1;
    vector<int> c(a.size() * 2 -1);
    int gas_count = a.size();
    for(int i=0; i<gas_count; i++)
    {
        c[i] = a[i] - g[i];
        if (i < gas_count - 1)
            c[i+gas_count] = a[i] - g[i];
    }
    int slow = 0;
    int fast = 0;
    int has_gas = 0;
    while(fast < c.size()){
        has_gas += c[fast];
        if (has_gas < 0) {
            slow = fast+1;
            has_gas = 0;
        }
        if (fast - slow +1 == gas_count) return slow;
        fast++;
    }
    return -1;
}
 Solution:
           1. 从左往右遍历,记住油量和最少的位置,从其下一个位置出发。
           2. 开辟一个长度为N的数组v,记录a[i]-g[i]。
              从后往前遍历数组v。
              如果v[i]小于零,将其与v[i-1]合并,因为此时i不可能作为起点。
              如果v[i]不小于零,记入sum,并记录该位置pos(有可能作为起点)。
              最后,如果v[0]大于等于零,说明整个路段已经没有负的v[i]。返回0即可。
              如果v[0]+sum>=0,有满足条件的加油站,返回pos。
              否则,返回-1。
           3. 开辟一个长度为2*N-1的数组v,记录a[i]-g[i](环转化为线性)。
              使用两个指针start和end。
              如果[start, end]区间和小于0,令start = end + 1并继续。
              直至找到长度为N的区间[start, end],并且区间和大于等于0。找到返回start。
           以上三种方案的时间复杂度皆为O(N)。
*/

int selectGasStation_1(const vector<int> &a, const vector<int> &g) {
    int N = a.size();
    int res = 0, min = a[0] - g[0], sum = min;
    for (int i = 1; i < N; ++i)
    {
        sum += a[i] - g[i];
        if (sum < min) {
            min = sum;
            res = i;
        }
    }
    return sum >= 0 ? (res + 1) % N : -1;
}

int selectGasStation_2(const vector<int> &a, const vector<int> &g) {
    int N = a.size();
    int v[N];
    for (int i = 0; i < a.size(); ++i)
        v[i] = a[i] - g[i];
    int sum = 0, pos = -1;
    for (int i = N-1; i > 0; --i)
    {
        if (v[i] >= 0) {
            sum += v[i];
            pos = i;
        } else {
            v[i-1] += v[i];
        }
    }
    if (v[0] >= 0) return 0;
    else if (v[0] + sum >= 0) return pos;
    else return -1;
}

int selectGasStation_3(const vector<int> &a, const vector<int> &g) {
    int N = a.size();
    int v[2 * N];
    for (int i = 0; i < N; ++i)
    {
        v[i] = a[i] - g[i];
        v[i+N] = a[i] - g[i];
    }
    int sum = 0;
    for (int start = 0, end = 0; start <= N && end < 2 * N; end++)
    {
        if (sum + v[end] < 0) {
            start = end + 1;
            sum = 0;
        } else {
            if (end - start == N - 1)
                return start;
            sum += v[end];
        }
    }
    return -1;
}

int selectGasStation(const vector<int> &a, const vector<int> &g)
{
vector<int> f;
if (a.size() <= 1)
return 0;
int size = a.size();
int i;
for (i = 0; i < size; ++ i)
f.push_back(a[i] - g[i]);
vector<int> result(2 * size + 1, -1);
result[0] = f[0];
int count = 0;
i = 1;
while (count < size && i <= 2*size)
{
if (result[i-1] >= 0) {
result[i] = result[i-1] + f[i%size];
count ++;
} else {
count = 0;
result[i] = f[i%size];
}
i ++;
}
if (count == size) {
return (i-1)%size;
}
return -1;
}
https://github.com/ralphite/ITint5-Answers-Java/blob/master/%E7%8E%AF%E5%BD%A2%E8%B7%AF%E5%8A%A0%E6%B2%B9.java
    public int selectGasStation(int[] a, int[] g) {
        if(a == null || g == null || a.length != g.length) return -1;
        int totalLeftGas = 0;
        int leftGas = 0;
        int startIndex = 0;
        for(int i = 0; i < a.length; i++) {
            totalLeftGas += a[i] - g[i];
            leftGas += a[i] - g[i];
            if(leftGas < 0) {
                startIndex = i + 1;
                leftGas = 0;
            }
        }
     
        return totalLeftGas >= 0 ? startIndex : -1;
    }
https://github.com/zdzapple/itint5/blob/master/%E7%8E%AF%E5%BD%A2%E8%B7%AF%E5%8A%A0%E6%B2%B9.cpp
/**
计算辅助数组f(i),其含义为从i点开始到达i+1时,汽车所剩的油。有:
             f(i) = g(i)-d(i)
若f(i)<0 则显然从i点出发不可能完成任务。


题目的O(n)解法依赖于以下结论,在面试的时间内,要迅速发现这个规律不是一件容易的事。
从任意一个点s开始扫描,如果到点e之前汽车没油了,那么说明从s到e-1中的所有点出发,都不可能完成任务。

设从s点到e时,油箱中剩下的油为t(s,e). 完成不了任务,说明t(s,e)<0
证明:
显然t(s,e) = f(s) + t(s+1,e)
若从s开始是合理的,则必有f(s)>=0
因此t(s+1,e) = t(s+1,e)-f(s)
若t(s,e)<0,那么必有t(s+1,e)<0,显然从s+1开始也完成不了任务。
由此可得结论。

对于任意一个O(n)复杂度的DP题目,都需要类似的结论做支持,即f(n)仅与一个变量n有关,如果和两个变量相关即f(s,n)那么复杂度必定不为O(n)。
这个题目中直观看到相关的变量有起始点和终止点2个,所以要通过分析,剔除和无关的变量,消除无需计算的部分,才能得出正确结果。
之后MS又加了一面,题目是有一个环路,中间有N个加油站,加油站里面的油是g1,g2...gn,加油站之间的距离是d1,d2...dn,问其中是否能找到一个加油站,使汽车从这个加油站出发,走完全程,这个题有一个O(n)的算法,最多扫描2N次元素,就是先从任意一个点a开始扫描,如果到点b和点b-1之前汽车没油了,那么说明点a到点b-1都不是可能的解,所以只要从点b开始继续扫就可以了,直到有一个点c,从c出发能够成功返回到c,那么c就是需要的解,如果再次扫描到了点a还没有找到解则说明无解.最坏情况就是c=a-1,那么需要扫描2N-1个加油站,当然可以在第一圈扫描时记载部分结果来使第二圈扫描更快.




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Orientation (1) Google APAC (1) Graph But No Graph (1) Graph Transpose (1) Graph Traversal (1) Graph-Coloring (1) Graph-Longest Path (1) Gray Code (1) HOJ (1) Hanoi (1) Hard Algorithm (1) How Hash (1) How to Test (1) Improve It (1) In Place (1) Inorder-Reverse Inorder Traverse Simultaneously (1) Interpolation search (1) Interview (1) Interview - Easy (1) Interview - Facebook (1) Isomorphic (1) JDK8 (1) K Dimensional Tree (1) Knapsack - Fractional (1) Knapsack - ZeroOnePack (1) Knight (1) Kosaraju’s algorithm (1) Kruskal (1) Kruskal MST (1) Kth Element (1) Least Common Ancestor (1) LeetCode - Binary Tree (1) LeetCode - Coding (1) LeetCode - Detail (1) LeetCode - Related (1) LeetCode Diffcult (1) Linked List Reverse (1) Linkedin (1) Linkedin Interview (1) Local MinMax (1) Logic Pattern (1) Longest Common Subsequence (1) Longest Common Substring (1) Longest Prefix Suffix(LPS) (1) Manhattan Distance (1) Map && Reverse Map (1) Math - Induction (1) Math-Multiply (1) Math-Sum Of Digits (1) Matrix - O(N+M) (1) Matrix BFS (1) Matrix Graph (1) Matrix Search (1) Matrix+DP (1) Matrix-Rotate (1) Max Min So Far (1) Median (1) Memory-Efficient (1) MinHash (1) MinMax Heap (1) Monotone Queue (1) Monto Carlo (1) Multi-Reverse (1) Multiple DFS (1) Multiple Tasks (1) Next Successor (1) Offline Algorithm (1) PAT (1) Partition (1) Path Finding (1) Patience Sort (1) Persistent (1) Pigeon Hole Principle (1) Power Set (1) Pratical Algorithm (1) Probabilistic Data Structure (1) Proof (1) Python (1) Queue & Stack (1) RSA (1) Ranking (1) Rddles (1) ReHash (1) Realtime (1) Recurrence Relation (1) Recursive DFS (1) Recursive to Iterative (1) Red-Black Tree (1) Region (1) Regular Expression (1) Resources (1) Reverse Inorder Traversal (1) Robin (1) Selection (1) Self Balancing BST (1) Similarity (1) Sort && Binary Search (1) String Algorithm. Symbol Table (1) String DP (1) String Distance (1) SubMatrix (1) Subsequence (1) System of Difference Constraints(差分约束系统) (1) TSP (1) Ternary Search Tree (1) Test (1) Thread (1) TimSort (1) Top-Down (1) Tournament (1) Tournament Tree (1) Transform Tree in Place (1) Tree Diameter (1) Tree Rotate (1) Trie + DFS (1) Trie and Heap (1) Trie vs Hash (1) Trie vs HashMap (1) Triplet (1) Two Data Structures (1) Two Stacks (1) USACO - Classical (1) USACO - Problems (1) UyHiP (1) Valid Tree (1) Vector (1) Wiggle Sort (1) Wikipedia (1) Yahoo Interview (1) ZOJ (1) baozitraining (1) codevs (1) cos126 (1) javabeat (1) jum (1) namic Programming (1) sqrt(N) (1) 两次dijkstra (1) 九度 (1) 二进制枚举 (1) 夹逼法 (1) 归一化 (1) 折半枚举 (1) 枚举 (1) 状态压缩DP (1) 男人八题 (1) 英雄会 (1) 逆向思维 (1)

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