[LeetCode] Jump Game II

This problem has a nice BFS structure. Let's illustrate it using the example

nums = [2, 3, 1, 1, 4]

in the problem statement. We are initially at position

0

. Then we can move at most

nums[0]

steps from it. So, after one move, we may reach

nums[1] = 3

or

nums[2] = 1

. So these nodes are reachable in

1

move. From these nodes, we can further move to

nums[3] = 1

and

nums[4] = 4

. Now you can see that the target

nums[4] = 4

is reachable in

2

moves.

Putting these into codes, we keep two pointers

start

and

end

that record the current range of the starting nodes. Each time after we make a move, update

start

to be

end + 1

and

end

to be the farthest index that can be reached in

1

move from the current

[start, end]

.

To get an accepted solution, it is important to handle all the edge cases. And the following codes handle all of them in a unified way without using the unclean

if

statements :-)

C++

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

Python

class Solution:
# @param {integer[]} nums
# @return {integer}
def jump(self, nums):
n, start, end, step = len(nums), 0, 0, 0
while end < n - 1:
step += 1
maxend = end + 1
for i in range(start, end + 1):
if i + nums[i] >= n - 1:
return step
maxend = max(maxend, i + nums[i])
start, end = end + 1, maxend
return step