inspired by the discussion in leetcode andhttp://tech-wonderland.net/blog/leetcode-jump-game-ii.html the keypoint of solving the problem by greedy approach is that we should keep the current maxium reachable distance, the next maxium reachable distance and also the steps needed to do it. w...
与[leetcode]Jump Game类似,不过数组不仅仅记录可达性,要记录最短路由跳数。思想一样。 例如[25000,24999,24998,24997,24996,24995,24994,24993...1]这个栗子,如果不记录最远可达(approach)的话,重复计算,势必超时 代码如下: 1publicclassSolution {2publicintjump(int[] a) {3if(a ==null|| a.length <...
jump length is 0, which makes it impossible to reach the last index. Solution 从nums数组末位开始向前遍历,用lastPos标记可达nums末位的最起始序号。 即lastPos 及之后元素均可通过一定步数到达last index. 1classSolution:2defcanJump(self, nums):3"""4:type nums: List[int]5:rtype: bool6"""7lastPo...
Jump Game 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. Determine if you are able to reach the last index. For example: A =[2,3,1,1,4], returntrue...
leetcode -- Jump Game II 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....