finish 724 solution

leetcode/0724.Find-Pivot-Index/724. Find Pivot Index.go

@@ -0,0 +1,23 @@
+package leetcode
+
+// 2 * leftSum + num[i] = sum
+// 时间: O(n)
+// 空间: O(1)
+func pivotIndex(nums []int) int {
+	if len(nums) <= 0 {
+		return -1
+	}
+
+	var sum, leftSum int
+	for _, num := range nums {
+		sum += num
+	}
+
+	for index, num := range nums {
+		if leftSum*2+num == sum {
+			return index
+		}
+		leftSum += num
+	}
+	return -1
+}

leetcode/0724.Find-Pivot-Index/724. Find Pivot Index_test.go

@@ -0,0 +1,52 @@
+package leetcode
+
+import "testing"
+
+func Test_pivotIndex(t *testing.T) {
+	type args struct {
+		nums []int
+	}
+	tests := []struct {
+		name string
+		args args
+		want int
+	}{
+		{
+			name: "test_case1",
+			args: args{[]int{1, 7, 3, 6, 5, 6}},
+			want: 3,
+		},
+		{
+			name: "test_case2",
+			args: args{[]int{1,2,3}},
+			want: -1,
+		},
+		{
+			name: "test_case3",
+			args: args{[]int{-1,-1,-1,-1,-1,-1}},
+			want: -1,
+		},
+		{
+			name: "test_case4",
+			args: args{[]int{-1,-1,-1,-1,-1,0}},
+			want: 2,
+		},
+		{
+			name: "test_case5",
+			args: args{[]int{1}},
+			want: 0,
+		},
+		{
+			name: "test_case5",
+			args: args{[]int{}},
+			want: -1,
+		},
+	}
+	for _, tt := range tests {
+		t.Run(tt.name, func(t *testing.T) {
+			if got := pivotIndex(tt.args.nums); got != tt.want {
+				t.Errorf("pivotIndex() = %v, want %v", got, tt.want)
+			}
+		})
+	}
+}

leetcode/0724.Find-Pivot-Index/README.md

@@ -0,0 +1,47 @@
+# [724. Find Pivot Index](https://leetcode.com/problems/find-pivot-index/)
+
+
+## 题目
+
+Given an array of integers nums, write a method that returns the "pivot" index of this array.
+
+We define the pivot index as the index where the sum of all the numbers to the left of the index is equal to the sum of all the numbers to the right of the index.
+
+If no such index exists, we should return -1. If there are multiple pivot indexes, you should return the left-most pivot index.
+
+ 
+
+**Example 1:**
+
+    Input: nums = [1,7,3,6,5,6]
+    Output: 3
+    Explanation:
+    The sum of the numbers to the left of index 3 (nums[3] = 6) is equal to the sum of numbers to the right of index 3.
+    Also, 3 is the first index where this occurs.
+
+**Example 2:**
+
+    Input: nums = [1,2,3]
+    Output: -1
+    Explanation:
+    There is no index that satisfies the conditions in the problem statement.
+
+**Constraints:**
+
+- The length of nums will be in the range [0, 10000].
+- Each element nums[i] will be an integer in the range [-1000, 1000].
+
+
+
+## 题目大意
+
+
+在数组中，找到一个数，使得它左边的数之和等于它的右边的数之和，如果存在，则返回这个数的下标索引，否作返回 -1
+
+
+## 解题思路
+
+- 这里面存在一个等式，只需要满足这个等式即可满足条件：leftSum + num[i] = sum - leftSum => 2 * leftSum + num[i] = sum
+- 题目提到如果存在多个索引，则返回最左边那个，因此从左开始求和，而不是从右边
+
+