Comments (2)
解题思路:
- 如果数组长度为1,返回唯一的数
- 定义两个指针分别指向数组的开头及结尾
- 检查数组是否翻转,如果没有返回最开始的那个数
- 当left小于right时,取中间作为mid进行二分查找
当mid右边值小于mid时返回nums[mid+1],
当mid左边值大于mid时返回nums[mid] - 上述两种都不符合时,如果left所在数小于mid所在数,则
left右移mid+1位置,否则将right左移至mid-1位置
/**
* @param {number[]} nums
* @return {number}
*/
var findMin = function (nums) {
if (nums.length < 2) {
return nums[0];
}
let left = 0;
let right = nums.length - 1;
if (nums[right] > nums[0]) {
return nums[0];
}
while (left < right) {
let mid = Math.floor(left + (right - left) / 2);
if (nums[mid] > nums[mid + 1]) {
return nums[mid + 1];
}
if (nums[mid - 1] > nums[mid]) {
return nums[mid];
}
if (nums[left] < nums[mid]) {
left = mid + 1;
} else {
right = mid - 1;
}
}
};
from like-algorithms.
/**
* @param {number[]} nums
* @return {number}
*/
var findMin = function (nums) {
const len = nums.length;
if (len == 1) return nums[0];
let left = 0;
let right = len - 1;
// 左边小于右边,表示未旋转,直接将最左返回
if (nums[left] < nums[right]) return nums[left];
while (left < right) {
const mid = Math.floor((left + right) / 2);
if (nums[mid] > nums[mid + 1]) return nums[mid + 1];
if (nums[mid] < nums[mid - 1]) return nums[mid];
if (nums[left] < nums[mid]) {
left = mid + 1;
} else {
right = mid - 1;
}
}
};
from like-algorithms.
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from like-algorithms.