四种方法解决JumpGame

出处：https://github.com/trekhleb/javascript-algorithms/tree/master/src/algorithms/uncategorized/jump-game

import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;/*** @description:* 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.** Input: [2,3,1,1,4]* Output: true* Explanation: Jump 1 step from index 0 to 1, then 3 steps to the last index.** We call a position in the array a "good" one if starting at that* position, we can reach the last index. Otherwise, that index* is called a "bad" one.*/
public class JumpGame {/*** 一、回溯法*/public static boolean jumpGameBacktracking(int[] numbers) {return helper(numbers, 0);}private static boolean helper(int[] numbers, int startIndex) {if (startIndex == numbers.length - 1) {// 退出条件,到达最后一个indexreturn true;}// check the longest jump we could make from current position// Math.min(jump is within array, jump goes beyond array)// 找到跳一步的极限，相当于剪枝，把超过步长的去掉int maxJumpLength = Math.min(numbers[startIndex], numbers.length - 1 - startIndex);// start to jump from startIndex,and see whether any jump is successful and has reached the end of the arrayfor (int i = maxJumpLength; i > 0; i--) {int nextIndex = startIndex + 1;// path.add(nextIndex); 应该helper函数无返回值的时候用path有用if (helper(numbers, startIndex + 1)) {return true;}// path.remove(path.size() - 1); 为了把走过的路回退，回到初始状态好进行下一次循环}return false;}/*** Top-Down自顶向下的动态规划，相当于优化过的回溯法* 题眼在于，只要我们确定了一个index是good or bad，那么结果就不会改变* 所以就可以把index的结果存起来不用每次都去重复计算* memorization：用一个数组记录每个index是good or bad，status：[GOOD, BAD, UNKNOWN]*/public static boolean jumpGameTopDown(int[] numbers) {int[] cellsGoodNess = new int[numbers.length];Arrays.fill(cellsGoodNess, 1);return topDownHelper(numbers, 0, new ArrayList<Integer>(), cellsGoodNess);}private static boolean topDownHelper(int[] numbers, int startIndex, List<Integer> path, int[] cellsGoodNess) {if (startIndex == numbers.length - 1) {return true;}int maxJumpLength = Math.min(numbers[startIndex], numbers.length - 1 - startIndex);for (int i = maxJumpLength; i > 0; i--) {int nextIndex = startIndex + i;if (cellsGoodNess[nextIndex] != 0) {path.add(nextIndex);if (topDownHelper(numbers, startIndex + 1, path, cellsGoodNess)) {return true;}path.remove(path.size() - 1);cellsGoodNess[nextIndex] = 0;}}return false;}/*** Bottom-Up 自底向上相比较自顶向下少了回溯的部分，省去了方法栈的多次调用* The observation to make here is that we only ever jump to the right.* This means that if we start from the right of the array,* every time we will query a position to our right, that position has already be determined as being GOOD or BAD.* This means we don't need to recurse anymore, as we will always hit the memo table.*/public static boolean jumpGameBottomUp(int[] numbers) {int[] cellsGoodness = new int[numbers.length];Arrays.fill(cellsGoodness, 0);// 将最后一个元素赋值为1cellsGoodness[numbers.length - 1] = 1;// 从右到左遍历数组for (int i = numbers.length - 2; i >= 0; i--) {int maxJumpLength = Math.min(numbers[i], numbers.length - 1 - i);for (int jumpLength = maxJumpLength; jumpLength > 0; jumpLength--) {int nextIndex = i + jumpLength;if (cellsGoodness[nextIndex] == 1) {// 如果后一个注定可以到达最后一个index，那么当前可以到nextIndex的i就也可以到达cellsGoodness[i] = 1;break;}}}// 最后的返回就是看第一个index的memo是否为true了return cellsGoodness[0] == 1;}/*** 贪心* Once we have our code in the bottom-up state, we can make one final, important observation.* From a given position, when we try to see if we can jump to a GOOD position, we only ever use one - the first one.* In other words, the left-most one.* If we keep track of this left-most GOOD position as a separate variable, we can avoid searching for it in the array.* Not only that, but we can stop using the array altogether.*/public static boolean jumpGameGreedy(int[] numbers) {int leftGoodPosition = numbers.length - 1;// 从右到左遍历for (int i = numbers.length - 2; i >= 0; i--) {// 如果可以从一个当前position到达good cell，那么当前position一定是goodint maxCurrentJumpLength = i + numbers[i];if (maxCurrentJumpLength >= leftGoodPosition) {// 当前位置经过当前最长步长能到达leftGoodPosition = i;}}// 判定最后的最后，leftGoodPosition是否为最左边的indexreturn leftGoodPosition == 0;}public static void main(String[] args) {int[] trueNumbers = new int[]{2,3,1,1,4};int[] falseNumbers = new int[]{3,2,1,0,4};System.out.println("jumpGameBacktracking true result:" + jumpGameBacktracking(trueNumbers));System.out.println("jumpGameBacktracking false result:" + jumpGameBacktracking(falseNumbers));System.out.println("jumpGameTopDown true result:" + jumpGameTopDown(trueNumbers));System.out.println("jumpGameTopDown false result:" + jumpGameTopDown(falseNumbers));System.out.println("jumpGameBottomUp true result:" + jumpGameBottomUp(trueNumbers));System.out.println("jumpGameBottomUp false result:" + jumpGameBottomUp(falseNumbers));System.out.println("jumpGameGreedy true result:" + jumpGameGreedy(trueNumbers));System.out.println("jumpGameGreedy false result:" + jumpGameGreedy(falseNumbers));}
}

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