LeetCode solution to a problem | 45. Jump Game II (jumping game Greedy C ++)

Title Description (Difficulty)

Original title link

algorithm

(greedy) $O (n)$

Each time you update the maximum distance, then every step of the jump can range as a range, whenever $i == end$ , then have to jump one step, if $end$ greater than or equal $size - 1$ , you can exit the for loop
Note: size = direct return 0:01

Time complexity is $O (n)$ , the spatial complexity is $O (1)$

C ++ code

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

        return steps;
    }
};

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