题目描述
给你一个下标从 0 开始、长度为 n
的整数数组 nums
,以及整数 indexDifference
和整数 valueDifference
。
你的任务是从范围 [0, n - 1]
内找出 2 个满足下述所有条件的下标 i
和 j
:
abs(i - j) >= indexDifference
且
abs(nums[i] - nums[j]) >= valueDifference
返回整数数组 answer
。如果存在满足题目要求的两个下标,则 answer = [i, j]
;否则,answer = [-1, -1]
。如果存在多组可供选择的下标对,只需要返回其中任意一组即可。
注意:i
和 j
可能 相等 。
示例 1:
输入:nums = [5,1,4,1], indexDifference = 2, valueDifference = 4
输出:[0,3]
解释:在示例中,可以选择 i = 0 和 j = 3 。
abs(0 - 3) >= 2 且 abs(nums[0] - nums[3]) >= 4 。
因此,[0,3] 是一个符合题目要求的答案。
[3,0] 也是符合题目要求的答案。
示例 2:
输入:nums = [2,1], indexDifference = 0, valueDifference = 0
输出:[0,0]
解释:
在示例中,可以选择 i = 0 和 j = 0 。
abs(0 - 0) >= 0 且 abs(nums[0] - nums[0]) >= 0 。
因此,[0,0] 是一个符合题目要求的答案。
[0,1]、[1,0] 和 [1,1] 也是符合题目要求的答案。
示例 3:
输入:nums = [1,2,3], indexDifference = 2, valueDifference = 4
输出:[-1,-1]
解释:在示例中,可以证明无法找出 2 个满足所有条件的下标。
因此,返回 [-1,-1] 。
提示:
1 <= n == nums.length <= 105
0 <= nums[i] <= 109
0 <= indexDifference <= 105
0 <= valueDifference <= 109
解法
方法一
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16 | class Solution:
def findIndices(
self, nums: List[int], indexDifference: int, valueDifference: int
) -> List[int]:
mi = mx = 0
for i in range(indexDifference, len(nums)):
j = i - indexDifference
if nums[j] < nums[mi]:
mi = j
if nums[j] > nums[mx]:
mx = j
if nums[i] - nums[mi] >= valueDifference:
return [mi, i]
if nums[mx] - nums[i] >= valueDifference:
return [mx, i]
return [-1, -1]
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22 | class Solution {
public int[] findIndices(int[] nums, int indexDifference, int valueDifference) {
int mi = 0;
int mx = 0;
for (int i = indexDifference; i < nums.length; ++i) {
int j = i - indexDifference;
if (nums[j] < nums[mi]) {
mi = j;
}
if (nums[j] > nums[mx]) {
mx = j;
}
if (nums[i] - nums[mi] >= valueDifference) {
return new int[] {mi, i};
}
if (nums[mx] - nums[i] >= valueDifference) {
return new int[] {mx, i};
}
}
return new int[] {-1, -1};
}
}
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22 | class Solution {
public:
vector<int> findIndices(vector<int>& nums, int indexDifference, int valueDifference) {
int mi = 0, mx = 0;
for (int i = indexDifference; i < nums.size(); ++i) {
int j = i - indexDifference;
if (nums[j] < nums[mi]) {
mi = j;
}
if (nums[j] > nums[mx]) {
mx = j;
}
if (nums[i] - nums[mi] >= valueDifference) {
return {mi, i};
}
if (nums[mx] - nums[i] >= valueDifference) {
return {mx, i};
}
}
return {-1, -1};
}
};
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19 | func findIndices(nums []int, indexDifference int, valueDifference int) []int {
mi, mx := 0, 0
for i := indexDifference; i < len(nums); i++ {
j := i - indexDifference
if nums[j] < nums[mi] {
mi = j
}
if nums[j] > nums[mx] {
mx = j
}
if nums[i]-nums[mi] >= valueDifference {
return []int{mi, i}
}
if nums[mx]-nums[i] >= valueDifference {
return []int{mx, i}
}
}
return []int{-1, -1}
}
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19 | function findIndices(nums: number[], indexDifference: number, valueDifference: number): number[] {
let [mi, mx] = [0, 0];
for (let i = indexDifference; i < nums.length; ++i) {
const j = i - indexDifference;
if (nums[j] < nums[mi]) {
mi = j;
}
if (nums[j] > nums[mx]) {
mx = j;
}
if (nums[i] - nums[mi] >= valueDifference) {
return [mi, i];
}
if (nums[mx] - nums[i] >= valueDifference) {
return [mx, i];
}
}
return [-1, -1];
}
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29 | impl Solution {
pub fn find_indices(nums: Vec<i32>, index_difference: i32, value_difference: i32) -> Vec<i32> {
let index_difference = index_difference as usize;
let mut mi = 0;
let mut mx = 0;
for i in index_difference..nums.len() {
let j = i - index_difference;
if nums[j] < nums[mi] {
mi = j;
}
if nums[j] > nums[mx] {
mx = j;
}
if nums[i] - nums[mi] >= value_difference {
return vec![mi as i32, i as i32];
}
if nums[mx] - nums[i] >= value_difference {
return vec![mx as i32, i as i32];
}
}
vec![-1, -1]
}
}
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