描述：通过最少的比较次数（1.5n），找到数组中最大的数和最小的数
思路1：每次取出两个数比较大小，较大者和最大值比得到最大值，较小者和最小值比得到最小值
/*
use least (1.5n) times compare to find the max and min value of an array
*/
public class FndMaxAndMin {
public class MaxAndMin {
int max;
int min;
MaxAndMin(int m, int n) {
max = m;
min = n;
}
}
MaxAndMin findMaxAndMin(int[] nums) {
int max = -1;
int min = -1;
int length = nums.length;
if (length == 0) {
return new MaxAndMin(max, min);
}
max = nums[0];
min = nums[0];
for (int i = 1; i < length / 2; i++) {
int tmpMax = nums[i] > nums[length - i] ? nums[i] : nums[length - i];
max = max > tmpMax ? max : tmpMax;
min = min < nums[i] + nums[length - i] - tmpMax ? min : nums[i] + nums[length - i] - tmpMax;
}
return new MaxAndMin(max, min);
}
}
