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.
For 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.)
Solution:
用动态规划DP的观点来实现。DP[i]代表到达i的最小跳数，可以证明DP是一个递增的数组，故可以快速剪枝。每次循环只需要尽量找到最小的DP[k]，使其满足k+A[k]>=n。
class Solution { public: int jump(int A[], int n) { if(n <= 1) return 0; int *dp = new int[n + 1]; for(int i=1;i<n;i++) dp[i] = INT_MAX; dp[0] = 0; for(int i = 1; i < n;i++) { for(int j = 0;j < i;j++) if(A[j] + j >= i) { if(dp[j] + 1 < dp[i]) { dp[i] = dp[j] + 1; break; } } } return dp[n - 1]; } };
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原文：http://www.cnblogs.com/changchengxiao/p/3594701.html