README_EN.md

33. Search in Rotated Sorted Array

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Description

There is an integer array nums sorted in ascending order (with distinct values).

Prior to being passed to your function, nums is possibly rotated at an unknown pivot index k (1 <= k < nums.length) such that the resulting array is [nums[k], nums[k+1], ..., nums[n-1], nums[0], nums[1], ..., nums[k-1]] (0-indexed). For example, [0,1,2,4,5,6,7] might be rotated at pivot index 3 and become [4,5,6,7,0,1,2].

Given the array nums after the possible rotation and an integer target, return the index of target if it is in nums, or -1 if it is not in nums.

You must write an algorithm with O(log n) runtime complexity.

 

Example 1:

Input: nums = [4,5,6,7,0,1,2], target = 0
Output: 4

Example 2:

Input: nums = [4,5,6,7,0,1,2], target = 3
Output: -1

Example 3:

Input: nums = [1], target = 0
Output: -1

 

Constraints:

• 1 <= nums.length <= 5000
• -104 <= nums[i] <= 104
• All values of nums are unique.
• nums is an ascending array that is possibly rotated.
• -104 <= target <= 104

Solutions

Binary search.

Python3

class Solution:
    def search(self, nums: List[int], target: int) -> int:
        n = len(nums)
        left, right = 0, n - 1
        while left < right:
            mid = (left + right) >> 1
            if nums[0] <= nums[mid]:
                if nums[0] <= target <= nums[mid]:
                    right = mid
                else:
                    left = mid + 1
            else:
                if nums[mid] < target <= nums[n - 1]:
                    left = mid + 1
                else:
                    right = mid
        return left if nums[left] == target else -1

Java

class Solution {
    public int search(int[] nums, int target) {
        int n = nums.length;
        int left = 0, right = n - 1;
        while (left < right) {
            int mid = (left + right) >> 1;
            if (nums[0] <= nums[mid]) {
                if (nums[0] <= target && target <= nums[mid]) {
                    right = mid;
                } else {
                    left = mid + 1;
                }
            } else {
                if (nums[mid] < target && target <= nums[n - 1]) {
                    left = mid + 1;
                } else {
                    right = mid;
                }
            }
        }
        return nums[left] == target ? left : -1;
    }
}

C++

class Solution {
public:
    int search(vector<int>& nums, int target) {
        int n = nums.size();
        int left = 0, right = n - 1;
        while (left < right) {
            int mid = (left + right) >> 1;
            if (nums[0] <= nums[mid]) {
                if (nums[0] <= target && target <= nums[mid])
                    right = mid;
                else
                    left = mid + 1;
            } else {
                if (nums[mid] < target && target <= nums[n - 1])
                    left = mid + 1;
                else
                    right = mid;
            }
        }
        return nums[left] == target ? left : -1;
    }
};

Go

func search(nums []int, target int) int {
	n := len(nums)
	left, right := 0, n-1
	for left < right {
		mid := (left + right) >> 1
		if nums[0] <= nums[mid] {
			if nums[0] <= target && target <= nums[mid] {
				right = mid
			} else {
				left = mid + 1
			}
		} else {
			if nums[mid] < target && target <= nums[n-1] {
				left = mid + 1
			} else {
				right = mid
			}
		}
	}
	if nums[left] == target {
		return left
	}
	return -1
}

JavaScript

/**
 * @param {number[]} nums
 * @param {number} target
 * @return {number}
 */
var search = function (nums, target) {
    const n = nums.length;
    let left = 0,
        right = n - 1;
    while (left < right) {
        const mid = (left + right) >> 1;
        if (nums[0] <= nums[mid]) {
            if (nums[0] <= target && target <= nums[mid]) {
                right = mid;
            } else {
                left = mid + 1;
            }
        } else {
            if (nums[mid] < target && target <= nums[n - 1]) {
                left = mid + 1;
            } else {
                right = mid;
            }
        }
    }
    return nums[left] == target ? left : -1;
};

Rust

impl Solution {
    pub fn search(nums: Vec<i32>, target: i32) -> i32 {
        let mut l = 0;
        let mut r = nums.len() - 1;
        while l <= r {
            let mid = l + r >> 1;
            if nums[mid] == target {
                return mid as i32;
            }

            if nums[l] <= nums[mid] {
                if target < nums[mid] && target >= nums[l] {
                    r = mid - 1;
                } else {
                    l = mid + 1;
                }
            } else {
                if target > nums[mid] && target <= nums[r] {
                    l = mid + 1;
                } else {
                    r = mid - 1;
                }
            }
        }
        -1
    }
}

TypeScript

function search(nums: number[], target: number): number {
    const n = nums.length;
    let left = 0,
        right = n - 1;
    while (left < right) {
        const mid = (left + right) >> 1;
        if (nums[0] <= nums[mid]) {
            if (nums[0] <= target && target <= nums[mid]) {
                right = mid;
            } else {
                left = mid + 1;
            }
        } else {
            if (nums[mid] < target && target <= nums[n - 1]) {
                left = mid + 1;
            } else {
                right = mid;
            }
        }
    }
    return nums[left] == target ? left : -1;
}

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