just find max height (height cut between height cut) after each cut
and find max width (width cut between width cut) after each cut
ans = max height * max width
time: O(nlogn)
space: O(1)
class Solution {
public int maxArea(int h, int w, int[] horizontalCuts, int[] verticalCuts) {
Arrays.sort(horizontalCuts);
Arrays.sort(verticalCuts);
int n = horizontalCuts.length;
int m = verticalCuts.length;
long maxH = Math.max(horizontalCuts[0], h - horizontalCuts[n-1]);
for (int i = 1; i < n; i++) {
maxH = Math.max(maxH, horizontalCuts[i] - horizontalCuts[i-1]);
}
long maxW = Math.max(verticalCuts[0], w - verticalCuts[m-1]);
for (int i = 1; i < m; i++) {
maxW = Math.max(maxW, verticalCuts[i] - verticalCuts[i-1]);
}
return (int)(maxH*maxW%1000000007); // notice here, (int) (product)
}
}
long maxH = horizontalCuts[0];
for (int i = 1; i < n; i++) {
maxH = Math.max(maxH, horizontalCuts[i] - horizontalCuts[i-1]);
}
maxH = Math.max(maxH, h - horizontalCuts[n-1]);
可以寫成下面這樣, 合併再一起去比較了 (先
long maxH = Math.max(horizontalCuts[0], h - horizontalCuts[n-1]);
