[LeetCode]Find Minimum in Rotated Sorted Array

思路1:最简单的思路莫过于排序然后取第一个元素,复杂度O(N*log(N))
代码1:

public class Solution {
    public int findMin(int[] num) {
        Arrays.sort(num);
        return num[0];
    }
}

思路2: 遍历,比较两段升序数字的第一个就可以了, 4 5 6 7 0 1 2复杂度O(N)
代码2:

    public int findMin(int[] num) {//解法1 遍历
        int min = num[0], currMin = num[0];
        int i = 0;
        while (++ i < num.length){
            currMin = Math.min(num[i],currMin);
            if(i + 1 < num.length && num[i + 1] < currMin){
                currMin = num[i + 1];
                break;
            }
        }
        min = Math.min(min, currMin);
        return min;
    }

    public int findMin(int[] num) {//解法3 二分
        int min = 0;
        int l = 0, r = num.length -1, mid;
        while (l < r){
            mid = l + (r - l) / 2;
            if(num[mid] > num[r]){
                l = mid + 1;
            }else {
                r = mid;
            }
        }
        min = num[l];
        return min;
    }

[LeetCode]Find Minimum in Rotated Sorted Array

原文：http://blog.csdn.net/youmengjiuzhuiba/article/details/44888087