45 jump game ii

Question

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.

Note:

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

Analysis

这一步能跳的范围是 i ~ i + nums[i], curFarthest 是之前所有点Max(i+nums[i]), 每次到达 i+nums[i] 发起另一个jump, curFarthest = i + nums[i]

Complexity

Time: O(n)

Solution

public class  {
    public int jump(int[] nums) {
        int curtStep = 0;
        int curtStepMaxReach = 0;
        int nextStepMaxReach = 0;
        for(int i = 0; i < nums.length - 1; i ++){
            nextStepMaxReach = Math.max(nextStepMaxReach, i + nums[i]);
            if(i == curtStepMaxReach){
                curtStep ++;
                curtStepMaxReach = nextStepMaxReach;
            }
        }
        return curtStep;
    }
}

45 jump game ii

Question

Analysis

Complexity

Solution

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