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1437.是否所有1都至少相隔k个元素.html
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1437.是否所有1都至少相隔k个元素.html
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<!DOCTYPE html>
<html lang="en">
<head>
<meta charset="UTF-8" />
<meta http-equiv="X-UA-Compatible" content="IE=edge" />
<meta name="viewport" content="width=device-width, initial-scale=1.0" />
<title>1437. 是否所有 1 都至少相隔 k 个元素</title>
</head>
<body>
<script>
// https://leetcode.cn/problems/check-if-all-1s-are-at-least-length-k-places-away/
// 给你一个由若干 0 和 1 组成的数组 nums 以及整数 k。如果所有 1 都至少相隔 k 个元素,则返回 True ;否则,返回 False 。
// 提示:
// 1 <= nums.length <= 10^5
// 0 <= k <= nums.length
// nums[i] 的值为 0 或 1
/**
* @param {number[]} nums
* @param {number} k
* @return {boolean}
*/
var kLengthApart = function (nums, k) {};
// --- answer-1 ---
// 遍历即可
// k大的情况 可以先判断last + k
var kLengthApart = function (nums = [], k) {
let last = -k - 1;
for (let i = 0; i < nums.length; i++) {
if (nums[i] === 1) {
if (i - last <= k) {
return false;
}
last = i;
}
}
return true;
};
// --- answer-1 ---
// --- answer-2 ---
// --- answer-2 ---
var nums = [1, 0, 0, 0, 1, 0, 0, 1],
k = 2;
var result = true;
// 解释:每个 1 都至少相隔 2 个元素。
// var nums = [1, 0, 0, 1, 0, 1],
// k = 2;
// var result = false;
// 解释:第二个 1 和第三个 1 之间只隔了 1 个元素。
var nums = [1, 1, 1, 1, 1],
k = 0;
var result = true;
var nums = [0, 1, 0, 1],
k = 1;
var result = true;
console.log('nums = ', nums);
console.log('k = ', k);
console.log('result = ', result);
console.log('kLengthApart = ', kLengthApart(nums, k));
</script>
</body>
</html>