代码实现：

方法一：递归+记忆化

int path;
int used[10000];

bool dfs(int *nums, int numsSize) {
    if (path == numsSize - 1) {
        return true;
    }
    for (int i = 1; i <= nums[path]; i++) {
        if (used[path + i]) {
            continue;
        }
        path += i;
        used[path] = 1;
        if (dfs(nums, numsSize)) {
            return true;
        }
        path -= i;
    }
    return false;
}

bool canJump(int *nums, int numsSize) {
    path = 0;
    memset(used, 0, sizeof(int) * numsSize);
    return dfs(nums, numsSize);
}

方法二：动态规划

bool canJump(int *nums, int numsSize) {
    bool dp[numsSize]; // 能否到达dp[i]
    memset(dp, false, sizeof(dp));
    dp[0] = true;
    for (int i = 1; i < numsSize; i++) {
        for (int j = 0; j < i; j++) {
            if (dp[j] && nums[j] + j >= i) {
                dp[i] = true;
                break;
            }
        }
    }
    return dp[numsSize - 1];
}

方法三：贪心

每次取最大跳跃步数（取最大覆盖范围）

bool canJump(int *nums, int numsSize) {
    if (numsSize == 1) {
        return true;
    }
    int cover = 0;
    for (int i = 0; i <= cover; i++) {
        cover = cover > i + nums[i] ? cover : i + nums[i];
        if (cover >= numsSize - 1) {
            return true;
        }
    }
    return false;
}