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.)
這是我的解法,思路類似於這個discuss,將此問題看做bfs.
int jump(vector<int>& nums)
{
int N = nums.size();
if (N <= 1) return 0;
// right is the right bound for this level
// cover is the right bound for next level
int right = 0 + nums[0], step = 1, cover = right;
for (int i = 1; i <= right; ++i)
{
if (right >= N - 1) return step;
cover = max(nums[i] + i, cover);
if (i == right)
{
right = cover;
++step;
}
}
return 0;
}