[LeetCode 55]Jump Game

题目链接：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], return true.A = [3,2,1,0,4], return false. * */public class JumpGame {//解法一//71 / 71 test cases passed.//Status: Accepted//Runtime: 236 ms//Submitted: 0 minutes ago//选中某点进行跳跃，然后在该点能到达的区间内选择跳得最远的点，进行下轮跳跃，如此循环//时间复杂度O(n) 空间复杂度O(1)public boolean canJump(int[] A) {//选中进行跳跃点的下标int choosedIndex = 0;while (choosedIndex < A.length) {//判断当前点是否能直接跳到 last indexif(choosedIndex + A[choosedIndex] >= A.length – 1) {return true;}//保存在区间[choosedIndex…choosedIndex+A[choosedIndex]]内跳得最远点的下标int max = choosedIndex;for (int i = choosedIndex + 1; i <= choosedIndex + A[choosedIndex]; i++) {if (A[i] + i > A[max] + max) {max =i;}}if (max == choosedIndex) {//无法跳到 last indexreturn false;}else {//选中跳得最远的点，进行下一轮跳跃choosedIndex = max;}}return false;}//解法二//71 / 71 test cases passed.//Status: Accepted//Runtime: 250 ms//Submitted: 0 minutes ago//时间复杂度O(n) 空间复杂度O(1)public boolean canJump1(int[] A) {int reach = 1;for (int i = 0; i < reach && reach < A.length; i++) {reach = Math.max(reach, 1 + i + A[i]);}return reach >= A.length;}public static void main(String[] args) {//System.out.println(canJump(new int[] {5,9,3,2,1,0,2,3,3,1,0,0}));}}

，想要成功，就一定要和成功的人在一起，不然反之