add: 跳跃游戏 II

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yi-ge 2020-05-04 13:43:56 +08:00
parent c5f7255d3a
commit b806ec7959
3 changed files with 30 additions and 0 deletions

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@ -235,6 +235,11 @@ LeetCode 与 LintCode 解题记录。此为个人练习仓库,代码中对重
- LeetCode 面试题42. 连续子数组的最大和 https://leetcode-cn.com/problems/lian-xu-zi-shu-zu-de-zui-da-he-lcof/ - LeetCode 面试题42. 连续子数组的最大和 https://leetcode-cn.com/problems/lian-xu-zi-shu-zu-de-zui-da-he-lcof/
- LintCode 41. 最大子数组 https://www.lintcode.com/problem/maximum-subarray/description - LintCode 41. 最大子数组 https://www.lintcode.com/problem/maximum-subarray/description
- [跳跃游戏II](src/array/jump-game-ii.js)
- LeetCode 45. 跳跃游戏 II https://leetcode-cn.com/problems/jump-game-ii/
- LintCode 117. 跳跃游戏 II https://www.lintcode.com/problem/jump-game-ii/description
## 栈 ## 栈
- [最大矩阵](src/stack/maximal-rectangle.js) - [最大矩阵](src/stack/maximal-rectangle.js)

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src/array/jump-game-ii.js Normal file
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/**
* @param {number[]} nums
* @return {number}
*/
export const jump = function (nums) {
let step = 0 // 跳跃步数
let maxPosition = 0 // 最大位置
let lastJumpStepMax = 0 // 最后一次跳跃最大能跳的步数
for (let n = 0, len = nums.length; n < len - 1; n++) {
maxPosition = Math.max(maxPosition, nums[n] + n)
if (lastJumpStepMax === n) { // 从当前位置开始,到当前能跳的位置这个区间,谁最大,就选谁,并跳一次
lastJumpStepMax = maxPosition
step++
}
if (lastJumpStepMax > len) return step
}
return step
}

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import { jump } from '../../src/array/jump-game-ii'
test('跳跃游戏II', () => {
expect(jump([2, 3, 1, 1, 4])).toBe(2)
})