leetcode Jump Game I II 待续 贪心看不懂啊！！！！

下面是这两个题的解法：

参考博客：http://blog.csdn.net/loverooney/article/details/38455475

自己写的第一题（TLE）：

#include<iostream>
#include<vector>

using namespace std;

bool canJump(vector<int>& nums) 
{
    int L = nums.size();
    bool *flag = new bool[L];
    for (int i = L - 2; i >= 0; i--)
    {
        int right = i + nums[i];
        if (right >= L - 1)
            flag[i] = true;
        else
        {
            flag[i] = false;
            for (int j = i + 1;j<L&& j < i + nums[i]; j++)
            {
                if (flag[j] == true)
                {
                    flag[i] = true;
                    break;
                }
            }
        }
    }
    return flag[0];
}

int main()
{
    vector<int> a = { 3, 2, 1, 0, 4 };
    cout << canJump(a) << endl;
}

贪心用step维护当前可到达的最远位置，每次的i必须在这个最远位置以内，假如不在就表明不能前进了return false。

代码：

#include<iostream>
#include<vector>

using namespace std;

bool canJump(vector<int>& nums) 
{
    int L = nums.size();
    int step = 0;
    for (int i = 0; i <= step; i++)
    {
        if (nums[i] + i > step)
            step = nums[i] + i;
        if (step >= L - 1)
            return true;
    }
    return false;
}

int main()
{
    int a[] = { 3, 2, 1, 0, 4 };
    cout << canJump(vector<int>(a, a+5)) << endl;
}

第二题：

BFS解法（MLE）：

#include<vector>
#include<iostream>
#include<queue>

using namespace std;

typedef struct Node
{
    int index;
    int jumpNo;
    Node(int index, int jumpNo) :index(index), jumpNo(jumpNo){};
}Node;

//    BFS
int jump(vector<int>& nums) 
{
    int i = 0; 
    int L = nums.size();
    queue<Node> que;
    que.push(Node(i,0));
    while (!que.empty())
    {
        Node now = que.front();
        que.pop();
        for (i = now.index + 1; i <= nums[now.index] + now.index; i++)
        {
            if (nums[i]+now.index >= L - 1)
            {
                return now.jumpNo+1;
            }
            else
                que.push(Node(i,now.jumpNo+1));
        }
    }
    return -1;
}

int main()
{
    vector<int> a = { 8, 3, 1, 1, 4 };
    cout << jump(a) << endl;
}

自己写的也是MLE：

#include<iostream>
#include<vector>
#include<queue>

using namespace std;

typedef struct Node
{
    int val;    //从该节点能调到哪儿
    int len;    //该节点在第len跳被访问
    int index;  //当前节点的下标
    Node(int val,int index,  int len) :val(val), index(index),len(len){};
    Node(){};
}Node;

int jump(vector<int>& nums) 
{
    int L = nums.size();
    Node* nodeArray = new Node[L];
    for (int i = 0; i < L; i++)
    {
        nodeArray[i].val = nums[i];
        nodeArray[i].index = i;
        nodeArray[i].len = 0;
    }
    //vector<bool> visited(L,false);
    queue<Node> visiteQueue;
    nodeArray[0].len = 0;
    visiteQueue.push(nodeArray[0]);
    while (!visiteQueue.empty())
    {
        Node temp = visiteQueue.front();
        int index = temp.index;
        int canStep = temp.val;
        for (int i = index + 1; i <= index + canStep; i++)
        {
            if (nodeArray[i].index+nodeArray[i].val>=L-1)
            {
                return temp.len + 2;
            }
            Node nextNode(nodeArray[i].val,nodeArray[i].index, temp.len + 1);
            visiteQueue.push(nextNode);
        }
        visiteQueue.pop();
    }
    return false;
}

int main()
{
    vector<int> a = { 2,1,3, 1, 1, 4,1,1,1,1};
    cout << jump(a) << endl;
}

内存不超的BFS：http://www.myext.cn/c/a_4468.html 

动态规划解法（TLE）：

#include<iostream>
#include<vector>

using namespace std;

int jump(vector<int>& nums) 
{
    int L = nums.size();
    int *dp = new int[L];
    dp[0] = 0;
    for (int i = 1; i < L; i++)
    {
        int min = INT_MAX;
        for (int j = 0; j < i; j++)
        {
            if (nums[j] + j >= i&&dp[j]+1<min)
            {
                min = dp[j] + 1;
            }
        }
        dp[i] = min;
    }
    for (int i = 0; i < L; i++)
        cout << dp[i] << endl;
    return dp[L - 1];
}

int main()
{
    vector<int> a = { 2, 3, 1, 1, 4 };
    cout << jump(a) << endl;
}

贪心解法：