45. Jump Game II

时间复杂度过高，o(n2)超时了：

class Solution {
public:
    int jump(vector<int>& nums) {
        int length = nums.size();
        if(length <= 0)
            return 0;
        vector<int> result(length);
        result[0] = 0;
        bool flag[length];
        flag[0] = true;
        for(int i = 1;i < length;i++)
            flag[i] = false;
        for(int i = 1;i < length;i++){
            int min_num = 0x7fffffff;
            for(int j = i-1;j >= 0;j--){
                if(flag[j] == false)
                    continue;
                if(flag[j] == true && nums[j] >= i-j){
                    if(result[j]+1 < min_num){
                        min_num = result[j] + 1;
                        flag[i] = true;
                    }
                }
            }
            result[i] = min_num;
        }
        return result[length-1];
    }
};

 把上面代码简化，依旧是超时

class Solution {
public:
    int jump(vector<int>& nums) {
        int length = nums.size();
        if(length <= 0)
            return 0;
        vector<int> result(length);
        result[0] = 0;
        for(int i = 1;i < length;i++)
            result[i] = -1;
        for(int i = 1;i < length;i++){
            int min_num = 0x7FFFFFFF;
            for(int j = i-1;j >= 0;j--){
                if(result[j] != -1 && nums[j] >= i-j){
                    result[i] = result[j] + 1;
                    if(result[i] < min_num)
                        min_num = result[i];
                }
            }
            if(min_num != 0x7FFFFFFF)
                result[i] = min_num;
            else
                result[i] = -1;
        }
        return result[length-1];
    }
};

 用贪心算法做，时间复杂度只有0(n)

class Solution {
public:
    int jump(vector<int>& nums) {
        int n = nums.size();
        if(n <= 0)
            return 0;
        int lastReach = 0;  
        int reach = 0;  
        int step = 0;  
        for(int i = 0; i < n && i <= reach; i++)  
        {  
            if(i > lastReach)  
            {  
                step++;  
                lastReach = reach;  
            }  
            reach = max(nums[i]+i, reach);  
        }  
        if(reach < n-1)  
            return 0;  
        else  
            return step;
    }
};

 推导过程：

两个讲解博客：

http://blog.csdn.net/cinderella_niu/article/details/42804559

http://www.cnblogs.com/boring09/p/4231771.html