[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.)

动态规划，注意剪枝，要不然要超时！

class Solution {
public:
    int jump(int A[], int n) {
        int *a = new int[n];
        for (int i = 0; i < n; ++i) {
            a[i] = INT_MAX;
        }
        a[0] = 0;
        for (int i = 1; i < n; ++i) {
            for (int j = 0; j < i; ++j) {
                if (j + A[j] >= i && a[i] > a[j] + 1) {
                    a[i] = a[j] + 1;
                    break;
                }
            }
        }
        return a[n-1];
    }
};

 果然HARD级别的题目有优于O(n^2)的算法，下面是O(n)的算法。

 res:目前为止的jump数

 curRch:从A[0]进行ret次jump之后达到的最大范围

 curMax:从0~i这i+1个A元素中能达到的最大范围

 当curRch < i，说明ret次jump已经不足以覆盖当前第i个元素，因此需要增加一次jump，使之达到记录的curMax。

class Solution {
public:
    int jump(int A[], int n) {
        int res = 0;
        int curRch = 0, curMax = 0;
        for (int i = 0; i < n; ++i) {
            if (curRch < i) {
                ++res;
                curRch = curMax;
            }
            curMax = max(curMax, i + A[i]);
        }
        return res;
    }
};