# LeetCode Jump Game II

LeetCode Jump Game II

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.)

Note:
You can assume that you can always reach the last index.

class Solution {
public:
int jump(vector<int>& nums) {
int n = nums.size();
if (n <= 1)return 0;
int maxJump = n - 1;
vector<int> dp(n, 0);
for (int i = n - 2; i >= 0; --i) {
int curMinJump = maxJump;
for (int j = i + 1; j <= nums[i] + i&&j < n; ++j) {
curMinJump = min(curMinJump, dp[j] + 1);
}
dp[i] = curMinJump;
}
return dp[0];
}
};


class Solution {
public:
int jump(vector<int>& nums) {
int n = nums.size();
if (n <= 1)return 0;
int maxJump = n - 1;
vector<int> dp(n, 0), pre(n, 0);
for (int i = n - 2; i >= 0; --i) {
int curMinJump = maxJump, k = 0;
for (int j = i + 1; j <= nums[i] + i&&j < n; ++j) {
if (dp[j] + 1 < curMinJump) {
curMinJump = dp[j] + 1;
k = j;
}
}
dp[i] = curMinJump;
pre[k] = i; // 从i跳到k能获得最小跳数
}
return dp[0];
}
};


class Solution {
public:
int jump(vector<int>& nums) {
int jumps = 0, curEnd = 0, curFarthest = 0;
for (int i = 0; i < nums.size() - 1; ++i) {
curFarthest = max(curFarthest, i + nums[i]);
if (i == curEnd) {
curEnd = curFarthest;
++jumps;
if (curEnd >= nums.size() - 1)break;
}
}
return jumps;
}
};