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.
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.)
Solution
class Solution:
# @param A, a list of integers
# @return an integer
# We use "last" to keep track of the maximum distance that has been reached
# by using the minimum steps "ret", whereas "curr" is the maximum distance
# that can be reached by using "ret+1" steps. Thus,curr = max(i+A[i]) where 0 <= i <= last.
def jump(self, A):
ret = 0
last = 0 #last is the steps that has been reached.
curr = 0 #max distance
for i in range(len(A)):
if i > last:
last = curr
ret += 1
curr = max(curr, i+A[i])
return ret
// version 1: Dynamic Programming
// 这个方法，复杂度是 O(n^2)，会超时，但是依然需要掌握。
public class Solution {
public int jump(int[] A) {
// state
int[] steps = new int[A.length];
// initialize
steps[0] = 0;
for (int i = 1; i < A.length; i++) {
steps[i] = Integer.MAX_VALUE;
}
// function
for (int i = 1; i < A.length; i++) {
for (int j = 0; j < i; j++) {
if (steps[j] != Integer.MAX_VALUE && j + A[j] >= i) {
steps[i] = Math.min(steps[i], steps[j] + 1);
}
}
}
// answer
return steps[A.length - 1];
}
}