Add Binary Search

Author: yudaer

docs/Leetcode_Solutions/Python/Summary/Binary Search.md

@@ -40,3 +40,92 @@ while (left<right)
 }
 酌情选用。
 ```
+
+
+
+@小白二号
+
+- 在一个有序的数组中寻找某个值
+
+```c++
+int bsearch(int* A, int n, int k) {
+    int left = -1, right = n;
+    while (right - left > 1) {
+        int mid = (left + right) / 2;
+        if (A[mid] == k) {
+            return mid;
+        } else if (A[mid] < k) {
+            left = mid;
+        } else {
+            right = mid;
+        }
+    }
+    return -1;
+}
+```
+
+- 满足judge(x)=true的最大值
+
+```c++
+bool judge(int x) {}
+int bsearch(int min, int max) {
+    int left = min, right = max;
+    //这里为什么要right++呢，因为我们求的是上界，二分区间中，left要满足f(x)，并且尽可能大
+    //需要要保证left在[min, max]遍历到
+    //终止条件是right - left > 1，所以right需要+1确保left能取到min
+    right++;
+    while (right - left > 1) {
+        int mid = (left + right) / 2;
+        if (judge(mid)) {
+            left = mid;
+        } else {
+            right = mid;
+        }
+    }
+    return left;
+}
+```
+
+- 满足judge(x)=true的最小值
+
+```c++
+bool judge(int x) {}
+int bsearch(int min, int max) {
+    int left = min, right = max;
+    //这里为什么要left--呢，因为我们求的是下界，二分区间中，right要满足f(x)，并且尽可能小
+    //需要要保证right在[min, max]遍历到
+    //终止条件是right - left > 1，所以left需要-1确保right能取到max
+    left--;
+    while (right - left > 1) {
+        int mid = (left + right) / 2;
+        if (judge(mid)) {
+            right = mid;
+        } else {
+            left = mid;
+        }
+    }
+    return right;
+}
+```
+
+- 浮点数，求f(x)在[min,max]的零点
+
+```c++
+const double EPS = 1e-8;
+double cal(double x) {} //假定是单调增
+//这里需要保证区间有零点
+double bsearch(double min, double max) {
+    double left = min, right = max;
+    while (right - left > EPS) {
+        double mid = (left + right) / 2;
+        if (A[mid] == 0) {
+            return mid;
+        } else if (cal(mid) < 0) {
+            left = mid;
+        } else {
+            right = mid;
+        }
+    }
+    return (left + right) / 2;
+}
+```