Swift-версия решения с анализом и результатами теста LeetCode.
/**
Calculate the area of intersection of two given rectilinear rectangles.
- Author:
Cong Liu <congliu0704 at gmail dot com>
- Returns:
The area of intersection of two given rectilinear rectangles.
- Parameters:
- K: The x coordinate of the lower left point of rectangle A
- L: The y coordinate of the lower left point of rectangle A
- M: The x coordinate of the upper right point of rectangle A
- N: The y coordinate of the upper right point of rectangle A
- P: The x coordinate of the lower left point of rectangle B
- Q: The y coordinate of the lower left point of rectangle B
- R: The x coordinate of the upper right point of rectangle B
- S: The y coordinate of the upper right point of rectangle B
- Assumptions:
All the eight given coordinates (K, L, M, N, P, Q, R and S) are integers
within the range [-2147483648...2147483647], that is, Int32-compatible.
K < M, L < N, P < R, Q < S
- Analysis:
The area of intersected is dyIntersected * dxIntersected.
To find out dyIntersected, consider how y coordinates of two rectangles relate
to each other, by moving rectangle A from above rectangle B down.
Case 1: when N > L >= S > Q, dyIntersected = 0
Case 2: when N >= S > L >= Q, dyIntersected = S - L
Case 3: when S > N > L >= Q, dyIntersected = N - L
Case 4: when S >= N >= Q > L, dyIntersected = N - Q
Case 5: when N > S > Q > L, dyIntersected = S - Q
Cases 2 and 3 can be merged as Case B:
when L >= Q, dyIntersected = min(N, S) - L
Cases 4 and 5 can be merged as Case C:
when Q > L, dyIntersected = min(N, S) - Q
Cases B and C can be merged as Case D:
when S > L , dyIntersected = min(N, S) - max(L, Q)
Likewise, x coordinates of two rectangles relate similarly to each other:
Case 1: when R > P >= M > K, dxIntersected = 0
Case 2: when M > P , dxIntersected = min(R, M) - max(P, K)
- Submission Date:
Sat 20 Jan 2018 CST at 23:28 pm
- Performance:
https://leetcode.com/problems/rectangle-area/description/
Status: Accepted
3081 / 3081 test cases passed.
Runtime: 78 ms
*/
class Solution {
public static func computeArea(_ K: Int, _ L: Int, _ M: Int, _ N: Int, _ P: Int, _ Q: Int, _ R: Int, _ S: Int) -> Int {
let areaA : Int = Int((M - K) * (N - L))
let areaB : Int = Int((R - P) * (S - Q))
var xIntersection : Int = 0
var yIntersection : Int = 0
var areaIntersection : Int = 0
if ((min(M, R) - max(K, P)) > 0) {
xIntersection = Int(min(M, R) - max(K, P))
}
if ((min(N, S) - max(L, Q)) > 0) {
yIntersection = Int(min(N, S) - max(L, Q))
}
if ((xIntersection == 0) || (yIntersection == 0)) {
areaIntersection = 0
} else {
areaIntersection = Int(xIntersection * yIntersection)
}
return (areaA + areaB - areaIntersection)
}
}
// A simple test
Solution.computeArea(-4, 1, 2, 6, 0, -1, 4, 3) // returns 42