LeetCode 45 Jump Game II (dp)

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本文介绍了一种高效算法来解决跳远游戏II问题，即如何在给定的非负整数数组中，从第一个索引开始，达到最后一个索引所需的最小跳跃次数。通过动态规划或改进的一次遍历方法，文章详细解释了如何实现这一目标。

Given an array of non-negative integers nums, 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.

You can assume that you can always reach the last index.

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 <= 104
0 <= nums[i] <= 1000

题目链接：https://leetcode.com/problems/jump-game-ii/

题目大意：求跳到最后最少要跳几次

题目分析：dp[i]表示到位置i最少跳的次数，维护一个当前可达的最大值 i + nums[i]即可

3ms，时间击败57.8%

class Solution {
    public int jump(int[] nums) {
        int n = nums.length;
        if (n < 2) {
            return 0;
        }
        int[] dp = new int[n];
        Arrays.fill(dp, n);
        int ma = 0;
        dp[0] = 0;
        for (int i = 0; i < nums.length; i++) {
            ma = Math.min(n - 1, Math.max(ma, i + nums[i]));
            dp[ma] = Math.min(dp[ma], dp[i] + 1);
        }
        return dp[n - 1];
    }
}

可以省去一维数组，思路类似，还是维护当前可达最大值，如果当前枚举值到了上一跳能到的最远处则再跳一次，跳的这一次就是从上一跳到当前这段里能跳到的最远处，枚举到n-2即可，因为n-1已经是终点了

1ms，时间击败99.57%

class Solution {
    public int jump(int[] nums) {
        int n = nums.length;
        if (n < 2) {
            return 0;
        }
        int ma = 0, curEnd = 0, ans = 0;
        for (int i = 0; i < n - 1; i++) {
            ma = Math.max(ma, i + nums[i]);
            if (i == curEnd) {
                curEnd = ma;
                ans++;
            }
        }
        return ans;
    }
}