Description#
You are given a 0-indexed array of integers nums of length n. You are initially positioned at nums[0].
Each element nums[i] represents the maximum length of a forward jump from index i. In other words, if you are at nums[i], you can jump to any nums[i + j] where:
0 <= j <= nums[i] andi + j < n
Return the minimum number of jumps to reach nums[n - 1]. The test cases are generated such that you can reach nums[n - 1].
Example 1:
Input: nums = [2,3,1,1,4]
Output: 2
Explanation: 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.
Example 2:
Input: nums = [2,3,0,1,4]
Output: 2
Constraints:
1 <= nums.length <= 1040 <= nums[i] <= 1000- It's guaranteed that you can reach
nums[n - 1].
Solutions#
Solution 1: Greedy Algorithm#
We can use a variable $mx$ to record the farthest position that can be reached from the current position, a variable $last$ to record the position of the last jump, and a variable $ans$ to record the number of jumps.
Next, we traverse each position $i$ in $[0,..n - 2]$. For each position $i$, we can calculate the farthest position that can be reached from the current position through $i + nums[i]$. We use $mx$ to record this farthest position, that is, $mx = max(mx, i + nums[i])$. Then, we check whether the current position has reached the boundary of the last jump, that is, $i = last$. If it has reached, then we need to make a jump, update $last$ to $mx$, and increase the number of jumps $ans$ by $1$.
Finally, we return the number of jumps $ans$.
The time complexity is $O(n)$, where $n$ is the length of the array. The space complexity is $O(1)$.
Similar problems:
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| class Solution:
def jump(self, nums: List[int]) -> int:
ans = mx = last = 0
for i, x in enumerate(nums[:-1]):
mx = max(mx, i + x)
if last == i:
ans += 1
last = mx
return ans
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| class Solution {
public int jump(int[] nums) {
int ans = 0, mx = 0, last = 0;
for (int i = 0; i < nums.length - 1; ++i) {
mx = Math.max(mx, i + nums[i]);
if (last == i) {
++ans;
last = mx;
}
}
return ans;
}
}
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| class Solution {
public:
int jump(vector<int>& nums) {
int ans = 0, mx = 0, last = 0;
for (int i = 0; i < nums.size() - 1; ++i) {
mx = max(mx, i + nums[i]);
if (last == i) {
++ans;
last = mx;
}
}
return ans;
}
};
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| func jump(nums []int) (ans int) {
mx, last := 0, 0
for i, x := range nums[:len(nums)-1] {
mx = max(mx, i+x)
if last == i {
ans++
last = mx
}
}
return
}
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| function jump(nums: number[]): number {
let [ans, mx, last] = [0, 0, 0];
for (let i = 0; i < nums.length - 1; ++i) {
mx = Math.max(mx, i + nums[i]);
if (last === i) {
++ans;
last = mx;
}
}
return ans;
}
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| impl Solution {
pub fn jump(nums: Vec<i32>) -> i32 {
let n = nums.len();
let mut dp = vec![i32::MAX; n];
dp[0] = 0;
for i in 0..n - 1 {
for j in 1..=nums[i] as usize {
if i + j >= n {
break;
}
dp[i + j] = dp[i + j].min(dp[i] + 1);
}
}
dp[n - 1]
}
}
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| public class Solution {
public int Jump(int[] nums) {
int ans = 0, mx = 0, last = 0;
for (int i = 0; i < nums.Length - 1; ++i) {
mx = Math.Max(mx, i + nums[i]);
if (last == i) {
++ans;
last = mx;
}
}
return ans;
}
}
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| #define min(a, b) a < b ? a : b
int jump(int* nums, int numsSize) {
int dp[numsSize];
for (int i = 0; i < numsSize; i++) {
dp[i] = numsSize;
}
dp[0] = 0;
for (int i = 0; i < numsSize - 1; i++) {
for (int j = i + 1; j < (min(i + nums[i] + 1, numsSize)); j++) {
dp[j] = min(dp[j], dp[i] + 1);
}
}
return dp[numsSize - 1];
}
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