154. Find Minimum in Rotated Sorted Array II

二刷

Runtime: 1 ms, faster than 67.57% of Java online submissions for Find Minimum in Rotated Sorted Array II.

Memory Usage: 43.3 MB, less than 61.79% of Java online submissions for Find Minimum in Rotated Sorted Array II.

class Solution {

    public int findMin(int[] nums) {

        if (nums == null) {

            throw new IllegalArgumentException();

        }

        int start = 0, end = nums.length - 1;

        while (start + 1 < end) {

            int mid = start + (end - start) / 2;

            if (nums[mid] < nums[end]) {

                end = mid;

            } else if (nums[mid] == nums[end]) {

                end--;

            } else {

                start = mid + 1;

            }

        }

        return Math.min(nums[start], nums[end]);

    }

}

一刷

Version #2 Better
只考虑是否舍弃后半段
若mid, end 相等有可能
1.[3, 3, 3, 3, 3, 3] All sorted
2.[3, 3, 3, 1, 3, 3] Left half sorted
3.[1, 2, 3, 3, 3, 3] Right half sorted
So the only thing we can do is reduce end pointer by 1
100.00 %
class Solution {
    public int findMin(int[] nums) {
        if (nums == null || nums.length == 0) {
            throw new IllegalArgumentException();
        }
        int start = 0;
        int end = nums.length - 1;
        while (start < end) { // [start, end]
            int mid = start + (end - start) / 2;
            if (nums[mid] == nums[end]) {
                end--;
            } else if (nums[mid] < nums[end]) {
                end = mid;
            } else { // nums[mid] > nums[end]
                start = mid + 1;
            }
        }
        return nums[start];
    }
}


Version #1
把 start, mid, end相等作为edge case
17.04 %
class Solution {
    public int findMin(int[] nums) {
        // [3, 3, 3, 1, 3, 3]
        // [1, 1, 1, 3, 3, 3]
       
        if (nums == null || nums.length == 0) {
            throw new IllegalArgumentException();
        }
        int start = 0;
        int end = nums.length - 1;
        while (start < end) { // [start, end]
            int mid = start + (end - start) / 2;
            if (nums[mid] == nums[end] && nums[start] == nums[mid]) {
                start++;
                end--;
            } else if (nums[mid] <= nums[end]) {
                end = mid;
            } else { // nums[mid] > nums[end]
                start = mid + 1;
            }
        }
        return nums[start];
    }
}