453. Minimum Moves to Equal Array Elements 难度:easy
2016-12-29 15:00
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题目:
Given a non-empty integer array of size n, find the minimum number of moves required to make all array elements equal, where a move is incrementing n - 1 elements by 1.
Example:
思路:
若使数组尽快满足题目要求,则需要每次给除去最大值的其余数字加1,可转换为每次将数组中的n-1个数字加1,相当于将剩余的一个数字减1。所以只需找到数组中的最小值m,计算m与数组中其他数字差的累计和即可。
程序:
class Solution {
public:
int minMoves(vector<int>& nums) {
sort(nums.begin(), nums.end());
int len = nums.size(), res = 0;
for (int i = 1; i < len; i++) {
res += nums[i] - nums[0];
}
return res;
}
};
Given a non-empty integer array of size n, find the minimum number of moves required to make all array elements equal, where a move is incrementing n - 1 elements by 1.
Example:
Input: [1,2,3] Output: 3 Explanation: Only three moves are needed (remember each move increments two elements): [1,2,3] => [2,3,3] => [3,4,3] => [4,4,4]
思路:
若使数组尽快满足题目要求,则需要每次给除去最大值的其余数字加1,可转换为每次将数组中的n-1个数字加1,相当于将剩余的一个数字减1。所以只需找到数组中的最小值m,计算m与数组中其他数字差的累计和即可。
程序:
class Solution {
public:
int minMoves(vector<int>& nums) {
sort(nums.begin(), nums.end());
int len = nums.size(), res = 0;
for (int i = 1; i < len; i++) {
res += nums[i] - nums[0];
}
return res;
}
};
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