Leetcode 45. Jump Game II

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;    }}