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:
1 2 3 4
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.
Note:
You can assume that you can always reach the last index.
Solution
Language: Java
从前往后 18.7% $O(n^2)$
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17
classSolution{ publicintjump(int[] nums){ if (nums == null || nums.length == 0) { return0; } // dp记录走到此处最小的jump数 int[] dp = newint[nums.length]; Arrays.fill(dp, Integer.MAX_VALUE); dp[0] = 0; for (int i = 0; i < dp.length; i++) { for (int j = i + 1; j <= Math.min(dp.length - 1, i + nums[i]); j++) { dp[j] = Math.min(dp[j], dp[i] + 1); } } return dp[nums.length - 1]; } }
贪心算法,每次都走最远的路 $O(n)$
1 2 3 4 5 6 7 8 9 10 11
publicintjump(int[] A){ int jumps = 0, curEnd = 0, curFarthest = 0; for (int i = 0; i < A.length - 1; i++) { curFarthest = Math.max(curFarthest, i + A[i]); if (i == curEnd) { jumps++; curEnd = curFarthest; } } return jumps; }