Leetcode 45. Jump Game II

最新推荐文章于 2022-07-10 16:38:50 发布

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最新推荐文章于 2022-07-10 16:38:50 发布

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分类专栏： Array Leetcode

本文链接：https://blog.youkuaiyun.com/u011094114/article/details/53948786

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本文介绍了一种基于广度优先搜索(BFS)的算法，用于解决数组中找到从起始位置到达最后一个位置所需的最少跳跃次数的问题。通过将数组中的每个数映射到一个层级，该算法能够有效地找出最短路径。

BFS solution for reference.

/**
 * Breadth first search o(n).
 * Map each number in the array to a level,
 * where numbers in level i are all the numbers that can be reached in i-1th jump.
 * e.g. 2, 3, 1, 1, 4
 * level 0: 2
 * level 1: 3, 1
 * level 2: 1, 4
 * The minimum jump is 2.
 */ 
public class Solution {
    public int jump(int[] nums) {
        if (nums.length < 2) return 0;
        int i = 1, lvl = 1, maxReach = nums[0], nextMax = 0;
        while (maxReach < nums.length-1) {
            for (; i<=maxReach; i++)
                // find the max reach of a level
                // there always exists a i+nums[i] >= maxReach (*)
                // so we can use nextMax to record the maxReach
                nextMax = Math.max(i+nums[i], nextMax); 
            // maxReach doesn't move forward after scanning a level
            // which means we can never jump to the end
            if (maxReach == nextMax) return Integer.MAX_VALUE;
            maxReach = nextMax;
            lvl++;
        }
        return lvl;
    }
}