LeetCode 题解 1537, 1541

Author: LjyYano

solution/1500-1599/1537.Get the Maximum Score/Solution.java

@@ -0,0 +1,44 @@
+class Solution {
+
+    final int MOD = (int) (1e9 + 7);
+
+    public int maxSum(int[] nums1, int[] nums2) {
+        Set<Integer> set1 = Arrays.stream(nums1).boxed().collect(Collectors.toSet());
+        Set<Integer> set2 = Arrays.stream(nums2).boxed().collect(Collectors.toSet());
+        set1.retainAll(set2);
+        if (set1.isEmpty()) {
+            return (int) (Math.max(sum(nums1, 0, nums1.length - 1), sum(nums2, 0, nums2.length - 1))
+                    % MOD);
+        }
+        long res = 0;
+        List<Integer> list = set1.stream().sorted(Comparator.naturalOrder())
+                .collect(Collectors.toList());
+        int start1 = 0, start2 = 0, end1 = 0, end2 = 0;
+        for (int common : list) {
+            // 2 个数组同时到达 set
+            while (nums1[end1] != common) {
+                end1++;
+            }
+            while (nums2[end2] != common) {
+                end2++;
+            }
+            // max
+            res += Math.max(sum(nums1, start1, end1), sum(nums2, start2, end2));
+            start1 = end1 + 1;
+            start2 = end2 + 1;
+        }
+
+        res += Math.max(sum(nums1, end1 + 1, nums1.length - 1),
+                sum(nums2, end2 + 1, nums2.length - 1));
+        res %= MOD;
+        return (int) res;
+    }
+
+    private long sum(int[] n, int l, int r) {
+        long res = 0;
+        while (l <= r) {
+            res += n[l++];
+        }
+        return res;
+    }
+}

solution/1500-1599/1541.Minimum Insertions to Balance a Parentheses String/Solution.java

@@ -0,0 +1,36 @@
+class Solution {
+     public int minInsertions(String s) {
+        int left = 0;
+        int res = 0;
+        char[] chars = s.toCharArray();
+        for (int i = 0; i < chars.length;) {
+            if (chars[i] == '(') {
+                left++;
+                i++;
+            } else {
+                // 连续2个 )
+                if (i < chars.length - 1 && chars[i + 1] == ')') {
+                    if (left > 0) {
+                        left--;
+                    } else {
+                        res++;
+                    }
+                    i += 2;
+                } else {
+                    if (left > 0) {
+                        left--;
+                        res++;
+                    } else {
+                        res += 2;
+                    }
+                    i++;
+                }
+            }
+
+        }
+        if (left > 0) {
+            res += 2 * left;
+        }
+        return res;
+    }
+}