标签：NPU ext range his bsp col let queue return
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.
Example:
Input: [2,3,1,1,4]
Output: 2
Explanation: The minimum number of jumps to reach the last index is 2.
Jump 1 step from index 0 to 1, then 3 steps to the last index.
Solution 1:
DP time: O(N^2) -> TLE
class Solution { public int jump(int[] nums) { if (nums == null || nums.length <= 1) { return 0; } int[] arr = new int[nums.length]; arr[0] = 0; for (int i = 1; i < nums.length; i++) { // initilize as -1, if set 0 leading to final res as 0 arr[i] = -1; for (int j = 0; j < i; j++) { if (arr[j] != -1 && j + nums[j] >= i) { if (arr[i] == -1 || arr[j] + 1 < arr[i]) { arr[i] = arr[j] + 1; } } } } return arr[nums.length - 1]; } }
Solution 2:
Greedy: O(N)
From LC, Let‘s say the range of the current jump is [curBegin, curEnd], curFarthest is the farthest point that all points in [curBegin, curEnd] can reach. Once the current point reaches curEnd, then trigger another jump, and set the new curEnd with curFarthest, then keep the above steps, as the following:
i == curEnd means you visited all the items on the current level. Incrementing jumps++ is like incrementing the level you are on. And curEnd = curFarthest is like getting the queue size (level size) for the next level you are traversing.
class Solution { public int jump(int[] nums) { if (nums == null || nums.length <= 1) { return 0; } int res = 0; int curMax = 0; int nextMax = 0; for(int i = 0; i < nums.length - 1; i++) { nextMax = Math.max(nextMax, i + nums[i]); // when last step, should not get this condition if (i == curMax) { res += 1; curMax = nextMax; } } return res; } }
标签：NPU ext range his bsp col let queue return
原文地址：https://www.cnblogs.com/xuanlu/p/12053671.html
