1064. Fixed Point 不动点

• 作者： 负雪明烛
• id： fuxuemingzhu
• 个人博客：https://fuxuemingzhu.cn/

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题目地址：https://leetcode-cn.com/problems/fixed-point/

题目描述

Given an array A of distinct integers sorted in ascending order, return the smallest index i that satisfies A[i] == i. Return -1 if no such i exists.

Example 1:

Input: [-10,-5,0,3,7]
Output: 3
Explanation:
For the given array, A[0] = -10, A[1] = -5, A[2] = 0, A[3] = 3, thus the output is 3.

Example 2:

Input: [0,2,5,8,17]
Output: 0
Explanation:
A[0] = 0, thus the output is 0.

Example 3:

Input: [-10,-5,3,4,7,9]
Output: -1
Explanation:
There is no such i that A[i] = i, thus the output is -1.

Note:

1. 1 <= A.length < 10^4
2. -10^9 <= A[i] <= 10^9

题目大意

给定已经按升序排列、由不同整数组成的数组 A，返回满足 A[i] == i 的最小索引 i。如果不存在这样的 i，返回 -1。

解题方法

暴力求解

从头遍历一遍即可。

C++代码如下：

class Solution {
public:
    int fixedPoint(vector<int>& A) {
        const int N = A.size();
        for (int i = 0; i < N; ++i) {
            if (A[i] == i)
                return i;
        }
        return -1;
    }
};

日期

2019 年 9 月 18 日 —— 今日又是九一八