题目：
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.)
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思路：
1）用一个dp数组来记录到达当前位置的最小步数
packagedp;publicclassJumpGameII {publicintjump(int[] nums) {intlen =nums.length;int[] dp =newint[len];for(inti = 0; i < len; ++i)
dp[i]=Integer.MAX_VALUE;
dp[0] = 0;//In order to save time, we don't visit the elements which already been visited.intstart = 1;for(inti = 0; i < len - 1; ++i) {
start=go(i, nums[i], start, dp);
}returndp[len-1];
}//Get next start position.privateintgo(intcurrent,intstep,intstart,int[] dp) {inti =start;for(; i <= current + step && i < dp.length; ++i) {if(dp[i] > dp[current] + 1)
dp[i]= dp[current] + 1;
}returni;
}publicstaticvoidmain(String[] args) {//TODO Auto-generated method stubint[] nums = { 2,3,1,1,4};
JumpGameII j=newJumpGameII();
System.out.println(j.jump(nums));
}
}
2）不用数组，就直接往前移动，第一次扫走一步能到的地方，第二次扫走二步能到的地方，在这个过程中，只要碰到了最后一个位置，就终止，证明是最小步数。
packagedp;publicclassJumpGameII {publicintjump(int[] nums) {intlen;if((len = nums.length) == 1)return0;intcount = 0;intstart = 0;intend = 0;while(end <len) {intmostRight = 0;++count;for(intj = start; j <= end; ++j) {if(j + nums[j] >= len - 1)returncount;if(j + nums[j] >mostRight)
mostRight= j +nums[j];
}
start= end + 1;
end=mostRight;
}returncount;
}publicstaticvoidmain(String[] args) {//TODO Auto-generated method stubint[] nums = { 2,3,1,1,4};
JumpGameII j=newJumpGameII();
System.out.println(j.jump(nums));
}
}