标签：fine task bsp 题解 ++ can tle put des
Leetcode - 624 - Maximum Distance in Arrays
Given m arrays, and each array is sorted in ascending order. Now you can pick up two integers from two different arrays (each array picks one) and calculate the distance. We define the distance between two integers a and b to be their absolute difference |a-b|. Your task is to find the maximum distance.
Example 1:
Input: [[1,2,3], [4,5], [1,2,3]] Output: 4 Explanation: One way to reach the maximum distance 4 is to pick 1 in the first or third array and pick 5 in the second array.
Note:
m arrays will be in the range of [2, 10000].m arrays will be in the range of [-10000, 10000]
题解：
使用 最大，次大，最小，次小 这四个元素来判断。 扫这个vector一次， O(n) 时间
其中 max_pt, min_pt 分别记录最大 最小取值的位置， 若两者一致，则从次大次小中来取得。
class Solution {
public:
int maxDistance(vector<vector<int>>& arrays) {
int ed, ans, min_f = 1000, min_s = 1000, max_f = -1000, max_s = -1000, min_pt = -1, max_pt = -1;
for(int i=0; i<arrays.size(); ++i){
if(arrays[i][0] < min_f){
min_s = min_f;
min_f = arrays[i][0];
min_pt = i;
}else if(arrays[i][0] < min_s){
min_s = arrays[i][0];
}
ed = arrays[i].size() - 1;
if(arrays[i][ed] > max_f){
max_s = max_f;
max_f = arrays[i][ed];
max_pt = i;
}else if(arrays[i][ed] > max_s){
max_s = arrays[i][ed];
}
}
if(max_pt == min_pt){
ans = max( max_f - min_s, max_s - min_f );
}else{
ans = max_f - min_f;
}
return ans;
}
};
Leetcode - 624 - Maximum Distance in Arrays
标签：fine task bsp 题解 ++ can tle put des
原文地址：http://www.cnblogs.com/zhang-yd/p/7067175.html
