1029. Two City Scheduling
https://leetcode.com/problems/two-city-scheduling/
use greedy
因為透過觀察,
Input: [[10,20],[30,200],[400,50],[30,20]]
cost[0] - cost[1] 的話, 會是 -10, -170, 350, 10
代表越小的結果, 應該派去 A, 會最好,
[[10,20],[30,200] 會選擇前者
本題說要一半的人去 A 一半的去 B, 所以透過排序 cost[0] - cost[1] , 由小到大
前一半的人去 A, 後一半的人去 B , 會是最佳解!
O(nlogn), O(1)
class Solution {
public int twoCitySchedCost(int[][] costs) {
Arrays.sort(costs, (a, b) -> {
return (a[0] - a[1]) - (b[0] - b[1]);
});
int res = 0;
for (int i = 0; i < costs.length; i++) {
if (i < costs.length/2) {
res += costs[i][0];
} else {
res += costs[i][1];
}
}
return res;
}
}
dp 解
dp[i][j]
represents the cost when considering first (i + j)
people in which i
people assigned to city A and j
people assigned to city B.
O(n^2), O(n^2), N = costs.length / 2
class Solution {
public int twoCitySchedCost(int[][] costs) {
int N = costs.length / 2;
int[][] dp = new int[N + 1][N + 1];
for (int i = 1; i <= N; i++) {
dp[i][0] = dp[i - 1][0] + costs[i - 1][0];
}
for (int j = 1; j <= N; j++) {
dp[0][j] = dp[0][j - 1] + costs[j - 1][1];
}
for (int i = 1; i <= N; i++) {
for (int j = 1; j <= N; j++) {
dp[i][j] = Math.min(dp[i - 1][j] + costs[i + j - 1][0], dp[i][j - 1] + costs[i + j - 1][1]);
}
}
return dp[N][N];
}
}
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