Question

256. Paint House

中文文档

Description

There is a row of n houses, where each house can be painted one of three colors: red, blue, or green. The cost of painting each house with a certain color is different. You have to paint all the houses such that no two adjacent houses have the same color.

The cost of painting each house with a certain color is represented by an n x 3 cost matrix costs.

• For example, costs[0][0] is the cost of painting house 0 with the color red; costs[1][2] is the cost of painting house 1 with color green, and so on...

Return _the minimum cost to paint all houses_.

 

Example 1:

Input: costs = [[17,2,17],[16,16,5],[14,3,19]]
Output: 10
Explanation: Paint house 0 into blue, paint house 1 into green, paint house 2 into blue.
Minimum cost: 2 + 5 + 3 = 10.

Example 2:

Input: costs = [[7,6,2]]
Output: 2

 

Constraints:

• costs.length == n
• costs[i].length == 3
• 1 <= n <= 100
• 1 <= costs[i][j] <= 20

Solutions

Solution 1

class Solution:
  def minCost(self, costs: List[List[int]]) -> int:
    a = b = c = 0
    for ca, cb, cc in costs:
      a, b, c = min(b, c) + ca, min(a, c) + cb, min(a, b) + cc
    return min(a, b, c)

class Solution {
  public int minCost(int[][] costs) {
    int r = 0, g = 0, b = 0;
    for (int[] cost : costs) {
      int _r = r, _g = g, _b = b;
      r = Math.min(_g, _b) + cost[0];
      g = Math.min(_r, _b) + cost[1];
      b = Math.min(_r, _g) + cost[2];
    }
    return Math.min(r, Math.min(g, b));
  }
}

class Solution {
public:
  int minCost(vector<vector<int>>& costs) {
    int r = 0, g = 0, b = 0;
    for (auto& cost : costs) {
      int _r = r, _g = g, _b = b;
      r = min(_g, _b) + cost[0];
      g = min(_r, _b) + cost[1];
      b = min(_r, _g) + cost[2];
    }
    return min(r, min(g, b));
  }
};

func minCost(costs [][]int) int {
  r, g, b := 0, 0, 0
  for _, cost := range costs {
    _r, _g, _b := r, g, b
    r = min(_g, _b) + cost[0]
    g = min(_r, _b) + cost[1]
    b = min(_r, _g) + cost[2]
  }
  return min(r, min(g, b))
}

/**
 * @param {number[][]} costs
 * @return {number}
 */
var minCost = function (costs) {
  let [a, b, c] = [0, 0, 0];
  for (let [ca, cb, cc] of costs) {
    [a, b, c] = [Math.min(b, c) + ca, Math.min(a, c) + cb, Math.min(a, b) + cc];
  }
  return Math.min(a, b, c);
};