Two ideas of Jump Game II (leetcode) DP: leetcodedp

Two ideas of Jump Game II (leetcode) DP: leetcodedp

The first approach is:

Dp (I): Minimum number of steps required to reach position I

Dp (I) must be incremental, so from j = A [I] (starting from the farthest position), update the array until dp (j + I) <= dp (I) + 1

If it is removed, it will

Int jump (int A [], int n) {int * dp = new int [n]; // the minimum number of steps required by dp [I] To I memset (dp, 0x3f, sizeof (int) * n); dp [0] = 0; for (int I = 0; I <n; I ++) {for (int j = A [I]; j> 0; j --) {if (j + I> = n-1) return min (dp [n-1], dp [I] + 1); if (dp [j + I] <= dp [I] + 1) break; dp [j + I] = dp [I] + 1 ;}} return dp [n-1];}

Another idea is:

MaxPos (I): The farthest position in step I

int jump(int A[], int n) {vector<int> maxPos;maxPos.push_back(0);maxPos.push_back(A[0]);if (n <= 1)return 0;if (A[0] >= n - 1)return 1;for(int index = 2; index < n; index++){int max = 0;for (int k = maxPos[index - 2] + 1; k <= maxPos[index - 1]; k++){if (max < A[k] + k){max = A[k] + k;if (max >= n - 1)return index;}}maxPos.push_back(max);}return n;}

Because we only use the index-1 and index-2 locations of maxPos, we can further optimize the space for O (1). We can use two variables to record the current maximum and the previous maximum positions.

The solution discussed in leetcode is as follows:

int jump(int A[], int n) {        int ret = 0;        int last = 0;        int curr = 0;        for (int i = 0; i < n; ++i) {            if (i > last) {                last = curr;                ++ret;            }            curr = max(curr, i+A[i]);        }        return ret;    }