题目一
给定一个只包含正数的数组arr，arr中任何一个子数组sub， 一定都可以算出(sub累加和 )* (sub中的最小值)是什么， 那么所有子数组中，这个值最大是多少？
public static int max2(int[] arr) {
int size = arr.length;
int[] sums = new int[size];
sums[0] = arr[0];
for (int i = 1; i < size; i++) {
sums[i] = sums[i - 1] + arr[i];
}
int max = Integer.MIN_VALUE;
Stack<Integer> stack = new Stack<Integer>();
for (int i = 0; i < size; i++) {
while (!stack.isEmpty() && arr[stack.peek()] >= arr[i]) {
int j = stack.pop();
max = Math.max(max, (stack.isEmpty() ? sums[i - 1] : (sums[i - 1] - sums[stack.peek()])) * arr[j]);
}
stack.push(i);
}
while (!stack.isEmpty()) {
int j = stack.pop();
max = Math.max(max, (stack.isEmpty() ? sums[size - 1] : (sums[size - 1] - sums[stack.peek()])) * arr[j]);
}
return max;
}
题目二
给定一个非负数组arr，代表直方图 返回直方图的最大长方形面积
public static int largestRectangleArea1(int[] height) {
if (height == null || height.length == 0) {
return 0;
}
int maxArea = 0;
Stack<Integer> stack = new Stack<Integer>();
for (int i = 0; i < height.length; i++) {
while (!stack.isEmpty() && height[i] <= height[stack.peek()]) {
int j = stack.pop();
int k = stack.isEmpty() ? -1 : stack.peek();
int curArea = (i - k - 1) * height[j];
maxArea = Math.max(maxArea, curArea);
}
stack.push(i);
}
while (!stack.isEmpty()) {
int j = stack.pop();
int k = stack.isEmpty() ? -1 : stack.peek();
int curArea = (height.length - k - 1) * height[j];
maxArea = Math.max(maxArea, curArea);
}
return maxArea;
}
public static int largestRectangleArea2(int[] height) {
if (height == null || height.length == 0) {
return 0;
}
int N = height.length;
int[] stack = new int[N];
int si = -1;
int maxArea = 0;
for (int i = 0; i < height.length; i++) {
while (si != -1 && height[i] <= height[stack[si]]) {
int j = stack[si--];
int k = si == -1 ? -1 : stack[si];
int curArea = (i - k - 1) * height[j];
maxArea = Math.max(maxArea, curArea);
}
stack[++si] = i;
}
while (si != -1) {
int j = stack[si--];
int k = si == -1 ? -1 : stack[si];
int curArea = (height.length - k - 1) * height[j];
maxArea = Math.max(maxArea, curArea);
}
return maxArea;
}
题目三
给定一个二维数组matrix，其中的值不是0就是1， 返回全部由1组成的最大子矩形，内部有多少个1
public static int maximalRectangle(char[][] map) {
if (map == null || map.length == 0 || map[0].length == 0) {
return 0;
}
int maxArea = 0;
int[] height = new int[map[0].length];
for (int i = 0; i < map.length; i++) {
for (int j = 0; j < map[0].length; j++) {
height[j] = map[i][j] == '0' ? 0 : height[j] + 1;
}
maxArea = Math.max(maxRecFromBottom(height), maxArea);
}
return maxArea;
}
// height是正方图数组
public static int maxRecFromBottom(int[] height) {
if (height == null || height.length == 0) {
return 0;
}
int maxArea = 0;
Stack<Integer> stack = new Stack<Integer>();
for (int i = 0; i < height.length; i++) {
while (!stack.isEmpty() && height[i] <= height[stack.peek()]) {
int j = stack.pop();
int k = stack.isEmpty() ? -1 : stack.peek();
int curArea = (i - k - 1) * height[j];
maxArea = Math.max(maxArea, curArea);
}
stack.push(i);
}
while (!stack.isEmpty()) {
int j = stack.pop();
int k = stack.isEmpty() ? -1 : stack.peek();
int curArea = (height.length - k - 1) * height[j];
maxArea = Math.max(maxArea, curArea);
}
return maxArea;
}
题目四
给定一个二维数组matrix，其中的值不是0就是1， 返回全部由1组成的子矩形数量
public static int numSubmat(int[][] mat) {
if (mat == null || mat.length == 0 || mat[0].length == 0) {
return 0;
}
int nums = 0;
int[] height = new int[mat[0].length];
for (int i = 0; i < mat.length; i++) {
for (int j = 0; j < mat[0].length; j++) {
height[j] = mat[i][j] == 0 ? 0 : height[j] + 1;
}
nums += countFromBottom(height);
}
return nums;
}
// 比如
// 1
// 1
// 1 1
// 1 1 1
// 1 1 1
// 1 1 1
//
// 2 .... 6 .... 9
// 如上图，假设在6位置，1的高度为6
// 在6位置的左边，离6位置最近、且小于高度6的位置是2，2位置的高度是3
// 在6位置的右边，离6位置最近、且小于高度6的位置是9，9位置的高度是4
// 此时我们求什么？
// 1) 求在3~8范围上，必须以高度6作为高的矩形，有几个？
// 2) 求在3~8范围上，必须以高度5作为高的矩形，有几个？
// 也就是说，<=4的高度，一律不求
// 那么，1) 求必须以位置6的高度6作为高的矩形，有几个？
// 3..3 3..4 3..5 3..6 3..7 3..8
// 4..4 4..5 4..6 4..7 4..8
// 5..5 5..6 5..7 5..8
// 6..6 6..7 6..8
// 7..7 7..8
// 8..8
// 这么多！= 21 = (9 - 2 - 1) * (9 - 2) / 2
// 这就是任何一个数字从栈里弹出的时候，计算矩形数量的方式
public static int countFromBottom(int[] height) {
if (height == null || height.length == 0) {
return 0;
}
int nums = 0;
int[] stack = new int[height.length];
int si = -1;
for (int i = 0; i < height.length; i++) {
while (si != -1 && height[stack[si]] >= height[i]) {
int cur = stack[si--];
if (height[cur] > height[i]) {
int left = si == -1 ? -1 : stack[si];
int n = i - left - 1;
int down = Math.max(left == -1 ? 0 : height[left], height[i]);
nums += (height[cur] - down) * num(n);
}
}
stack[++si] = i;
}
while (si != -1) {
int cur = stack[si--];
int left = si == -1 ? -1 : stack[si];
int n = height.length - left - 1;
int down = left == -1 ? 0 : height[left];
nums += (height[cur] - down) * num(n);
}
return nums;
}
public static int num(int n) {
return ((n * (1 + n)) >> 1);
}
题目五
给定一个数组arr，返回所有子数组最小值的累加和
leetcode.com/problems/su…
public static int sumSubarrayMins(int[] arr) {
int N = arr.length;
int[] stack = new int[N];
int[] left = new int[N];
int[] right = new int[N];
int si = -1;
for (int i = 0; i < N; i++) {
// 大于号表示：相等的时候，右边界可以算对，左边界是上一个相等的数
while (si != -1 && arr[stack[si]] > arr[i]) {
int cur = stack[si--];
int l = si == -1 ? -1 : stack[si];
left[cur] = l;
right[cur] = i;
}
stack[++si] = i;
}
while (si != -1) {
int cur = stack[si--];
int l = si == -1 ? -1 : stack[si];
left[cur] = l;
right[cur] = N;
}
long ans = 0;
for (int i = 0; i < arr.length; i++) {
long start = i - left[i];
long end = right[i] - i;
ans += start * end * (long) arr[i];
ans %= 1000000007;
}
return (int) ans;
}
