标签:solution eps || length lintcode class for step integer
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
Given array A = [2,3,1,1,4]
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.)
public class Solution {
/**
* @param A: A list of integers
* @return: An integer
*/
public int jump(int[] A) {
// 最少要多少步能跳到最后一个
if(A == null || A.length <= 0) return 0;
int[] dp = new int[A.length];
dp[0] = 0;
for(int i = 1; i < A.length; i++) {
dp[i] = Integer.MAX_VALUE;
for(int j = 0; j < i; j++) {
if(j + A[j] >= i) {
dp[i] = Math.min(dp[i], dp[j]+1);
}
}
}
return dp[A.length-1];
}
}
标签:solution eps || length lintcode class for step integer
原文地址:https://www.cnblogs.com/lawrenceSeattle/p/10276304.html