[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 :-)

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C++

 1 class Solution {
 2 public:
 3     int jump(vector<int>& nums) {
 4         int n = nums.size(), step = 0, start = 0, end = 0;
 5         while (end < n - 1) {
 6             step++; 
 7             int maxend = end + 1;
 8             for (int i = start; i <= end; i++) {
 9                 if (i + nums[i] >= n - 1) return step;
10                 maxend = max(maxend, i + nums[i]);
11             }
12             start = end + 1;
13             end = maxend;
14         }
15         return step;
16     }
17 };

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Python

 1 class Solution:
 2     # @param {integer[]} nums
 3     # @return {integer}
 4     def jump(self, nums):
 5         n, start, end, step = len(nums), 0, 0, 0
 6         while end < n - 1:
 7             step += 1
 8             maxend = end + 1
 9             for i in range(start, end + 1):
10                 if i + nums[i] >= n - 1:
11                     return step
12                 maxend = max(maxend, i + nums[i])
13             start, end = end + 1, maxend
14         return step