528. Random Pick with Weight
class Solution:
def __init__(self, w: List[int]):
self.running_sums = []
running_sum = 0
for weight in w:
running_sum += weight
self.running_sums.append(running_sum)
self.total_sum = running_sum
def pickIndex(self) -> int:
target = random.randint(1, self.total_sum)
low, high = 0, len(self.running_sums)
while low < high:
mid = low + (high - low) // 2
if target > self.running_sums[mid]:
low = mid + 1
else:
high = mid
return low
497. Random Point in Non-overlapping Rectangles
class Solution:
def __init__(self, rects: List[List[int]]):
self.rects = rects
self.areas = []
self.total_area = 0
for rect in rects:
x1, y1, x2, y2 = rect
area = (x2 - x1 + 1) * (y2 - y1 + 1)
# x1 = 1，x2 = 3，矩形的宽度应该是包含从 x1 到 x2 的所有点，即 1、2、3 这三个 x 坐标。所以宽度是 3 - 1 + 1 = 3
self.total_area += area
self.areas.append(self.total_area)
def pick(self) -> List[int]:
target = random.randint(1, self.total_area)
idx = bisect.bisect_left(self.areas, target)
x1, y1, x2, y2 = self.rects[idx]
x = random.randint(x1, x2)
y = random.randint(y1, y2)
return [x,y]
'''
计算每个矩形的面积：每个矩形的面积越大，被选择到的概率也应该越大。因此，我们首先要计算出每个矩形的面积，并且维护一个前缀和数组 areas，用来存储这些矩形面积的累加值。
根据前缀和数组来选择矩形：我们可以使用一个随机数生成器来生成一个随机的整数，然后通过二分查找在前缀和数组中找到对应的矩形。
在选中的矩形内随机选择一个点：一旦选定了一个矩形，我们可以在这个矩形的边界内随机选择一个点。
'''
# Your Solution object will be instantiated and called as such:
# obj = Solution(rects)
# param_1 = obj.pick()
33. Search in Rotated Sorted Array
class Solution:
def search(self, nums: List[int], target: int) -> int:
left, right = 0, len(nums) - 1
while left <= right:
mid = (left + right) // 2
if nums[mid] == target:
return mid
if nums[left] <= nums[mid]:
if nums[left] <= target < nums[mid]:
right = mid - 1
else:
left = mid + 1
if nums[mid] <= nums[right]:
if nums[mid] < target <= nums[right]:
left = mid + 1
else:
right = mid - 1
return -1
81. Search in Rotated Sorted Array II
class Solution:
def search(self, nums: List[int], target: int) -> bool:
low, high = 0, len(nums) - 1
while low <= high:
mid = (low + high) // 2
if nums[mid] == target:
return True
if nums[low] == nums[mid]:
low += 1
continue
if nums[low] <= nums[mid]:
if nums[low] <= target <= nums[mid]:
high = mid - 1
else:
low = mid + 1
else:
if nums[mid] <= target <= nums[high]:
low = mid + 1
else:
high = mid - 1
return False
375. Guess Number Higher or Lower II(dp
class Solution {
int[][] best;
public int getMoneyAmount(int n) {
best = new int[n + 1][n + 1];
for (int i = 0; i < n + 1; i++) {
Arrays.fill(best[i], -1);
}
return dp(1, n);
}
private int dp(int lo, int hi) {
if (lo >= hi) {
return 0;
} else {
if (best[lo][hi] != -1) {
return best[lo][hi];
}
//the idea is to compute the maximum amount of money I need to pay with my guesses ranging from lo to hi.
//Since I want to minimize my payoff, I'll choose the starting point [lo,hi] that has the minimum payoff.
//But for each starting point, I'll choose the maximum amount of money I pay, since I need the worst case scenario.
int res = Integer.MAX_VALUE;
for (int i = lo; i <= hi; i++) {
int op1 = 0; //guess = i, i was correct, no cost incurred
int op2 = i + dp(lo, i - 1);//guess = i = wrong, go lo
int op3 = i + dp(i + 1, hi);//guess = i = wrong, go hi
int out = Math.max(op1, Math.max(op2, op3));
res = Math.min(out, res);
}
best[lo][hi] = res;
return res;
}
}
}
374. Guess Number Higher or Lower
public class Solution extends GuessGame {
public int guessNumber(int n) {
int left = 1, right = n;
while(left <= right) {
int mid = left + (right - left) / 2;
int num = guess(mid);
if(num == 0) {
return mid;
} else if(num == -1) {
right = mid - 1;
} else if(num == 1) {
left = mid + 1;
}
}
return -1;
}
}
35. Search Insert Position
class Solution {
public int searchInsert(int[] nums, int target) {
int left = 0, right = nums.length - 1;
while(left <= right) {
int mid = left + (right - left) / 2;
if(nums[mid] == target) {
return mid;
} else if(nums[mid] < target) {
left = mid + 1;
} else {
right = mid - 1;
}
}
return left;
}
}
278. First Bad Version
public class Solution extends VersionControl {
public int firstBadVersion(int n) {
int left = 1, right = n;
while(left <= right) {
int mid = left + (right - left) / 2;
if(isBadVersion(mid)) {
right = mid - 1;
} else {
left = mid + 1;
}
}
return left;
}
}
367. Valid Perfect Square
class Solution {
public boolean isPerfectSquare(int num) {
int left = 1, right = num;
while(left <= right) {
int mid = left + (right - left) / 2;
// mid * mid 会越界
int div = num / mid;
if(div == mid) {
if(num % mid == 0) {
return true;
}
left = mid + 1;
} else if(div < mid) {
right = mid - 1;
} else {
left = mid + 1;
}
}
return false;
}
}
69. Sqrt(x)
class Solution {
public int mySqrt(int x) {
int l = 1, r = x;
while(l <= r) {
int mid = l + (r - l) / 2;
if((long) mid * mid == (long) x ) {
return mid;
} else if((long) mid * mid < (long) x) {
l = mid + 1;
} else {
r = mid - 1;
}
}
return Math.round(r);
}
}
441. Arranging Coins
class Solution {
public int arrangeCoins(int n) {
if(n < 0) {
throw new IllegalArgumentException("only positve number allowed");
}
if(n <= 1) {
return n;
}
if(n <= 3) {
return n == 3 ? 2 : 1;
}
long left = 0, right = n / 2;
while(left <= right) {
long mid = left + (right - left) / 2;
long sum = mid * (mid + 1) / 2;
if(sum == n) {
return (int)mid;
} else if(sum < n) {
left = mid + 1;
} else {
right = mid - 1;
}
}
return (int)right;
}
}
34. Find First and Last Position of Element in Sorted Array
class Solution {
public int[] searchRange(int[] nums, int target) {
int left = 0, right = nums.length - 1;
int[] ans = {-1, -1};
while(left <= right) {
int mid = left + (right - left) / 2;
if(nums[mid] == target) {
ans[0] = mid;
right = mid - 1;
} else if(nums[mid] > target) {
right = mid - 1;
} else {
left = mid + 1;
}
}
left = 0;
right = nums.length - 1;
while(left <= right) {
int mid = left + right >> 1;
if(target == nums[mid]) {
ans[1] = mid;
left = mid + 1;
} else if(target < nums[mid]) {
right = mid - 1;
} else {
left = mid + 1;
}
}
return ans;
}
}
136. Single Number
class Solution {
public int singleNumber(int[] nums) {
for(int i = 1; i < nums.length; i++) {
nums[0] ^= nums[i];
}
return nums[0];
}
}
//异或运算：0 XOR 任何数等于任何数， 任何相同的二进制数进行XOR都等于0；
137. Single Number II
class Solution {
public int singleNumber(int[] nums) {
Map<Integer, Integer> map = new HashMap();
for(int num : nums) {
map.put(num, map.getOrDefault(num, 0) + 1);
}
for(Map.Entry<Integer, Integer> entry : map.entrySet()) {
if(entry.getValue() == 1) {
return entry.getKey();
}
}
return -1;
}
}
class Solution {
public int singleNumber(int[] nums) {
Arrays.sort(nums);
int n = nums.length - 1;
if (n == 0 || nums[0] != nums[1]) return nums[0];
if (nums[n] != nums[n - 1]) return nums[n];
for (int i = 1; i < n; i++) {
if (nums[i] != nums[i - 1] && nums[i] != nums[i + 1]) return nums[i];
}
return 0;
}
}
540. Single Element in a Sorted Array
class Solution {
public int singleNonDuplicate(int[] nums) {
int left = 0, right = nums.length - 1;
while(left < right) {
int mid = left + (right - left) / 2;
if(mid % 2 == 1) {
mid--;
}
if(nums[mid] == nums[mid + 1]) {
left = mid + 2;
} else {
right = mid;
}
}
return nums[left];
}
}
275. H-Index II
class Solution {
public int hIndex(int[] citations) {
int left = 0, right = citations.length - 1, res = 0;
while(left <= right) {
int mid = (left + right) >> 1;
int len = citations.length - mid;
if(citations[mid] >= len) {
res = len;
right = mid - 1;
} else {
left = mid + 1;
}
}
return res;
}
}
1539. Kth Missing Positive Number
class Solution {
public int findKthPositive(int[] arr, int k) {
int left = 0, right = arr.length - 1;
while(left <= right) {
int mid = left + right >> 1;
if(arr[mid] - mid - 1 < k) {
left = mid + 1;
} else {
right = mid - 1;
}
}
return left + k;
}
}
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