Leetcode 每日一题C 语言版 -- 45 jump game ii

小武~2025-12-07 22:20

Leetcode 每日一题C 语言版 -- 45 jump game ii

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问题描述

https://leetcode.com/problems/jump-game-ii/?envType=study-plan-v2\&envId=top-interview-150

复制代码

You are given a 0-indexed array of integers nums of length n. You are initially positioned at index 0.

Each element nums[i] represents the maximum length of a forward jump from index i. In other words, if you are at index i, you can jump to any index (i + j) where:

0 <= j <= nums[i] and
i + j < n
Return the minimum number of jumps to reach index n - 1. The test cases are generated such that you can reach index n - 1.

 

Example 1:

Input: nums = [2,3,1,1,4]
Output: 2
Explanation: 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.
Example 2:

Input: nums = [2,3,0,1,4]
Output: 2
 

Constraints:

1 <= nums.length <= 104
0 <= nums[i] <= 1000
It's guaranteed that you can reach nums[n - 1].

解法一

greddy algo.

记录最远能跳到的位置，然后跳到这个位置p1。

遍例当前位置~这个最远位置，得到一个更新后的最远位置p2。

遍历到p1 时, 一定可以跳到p2, 将位置更新到p2

c 复制代码

int jump(int* nums, int numsSize) {
    uint32_t farthest = 0, current = 0;
    uint32_t index = 0;
    int jumps = 0;

    for (index = 0; index < numsSize - 1; index++) {
        if (index + nums[index] > farthest)
            farthest = index + nums[index];
        if (index == current) {
            current = farthest;
            jumps++;
        }
    }
    return jumps;
}