[LeetCode] 45. Jump Game II

45. Jump Game II

Description

Given an array of non-negative integers, you are initially positioned at the first index of the array.

Each element in the array represents your maximum jump length at that position.

Your goal is to reach the last index in the minimum number of jumps.

For example:
Given array A = [2,3,1,1,4]

The minimum number of jumps to reach the last index is 2. (Jump 1 step from index 0 to 1, then 3 steps to the last index.)

Note:
You can assume that you can always reach the last index.

Algorithm

// beats 76.54% of cpp on 2017-02-12

class Solution {
public:
    int jump(vector<int>& nums) {
        vector<int> mins(nums.size());
        int n = nums.size();
        if (n <= 1) return 0;
        for (int i = 0; i < n; i++) {
            mins[i] = -1;
        }
        mins[0] = 0;
        for (int i = 0; i < n; i++) {
            for (int j = nums[i]; j >= 1; j--) if (i + j < n) {
                if (mins[i+j] == -1) mins[i+j] = mins[i] + 1;
                else mins[i+j] = min(mins[i+j], mins[i] + 1);
                // (1) if (i + j == n - 1) return mins[n-1];
                // (2) if (i + j + nums[i+j] >= n-1) return mins[i+j] + 1;
            }
        }
        return mins[n-1];
    }
};

把注释（1）代码加进去来加速，还是超时，因为最卡时间的那个例子是与数据大小的平方成正比的。但是再把注释（2）的那句代码加上就把这种情况避免掉了。有本事就会再来一层（笑哭）不过那样的话数据规模就会更大了。